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Find the number of real solutions of equation : $sin (3theta) = 4sinthetasin(2theta) sin(4theta)$ in $theta in (0, pi)$.
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My try : Sorry , but I can't figure out what is the first step to do in this question . But still I tried and converted $sin(3theta)$ into $sin (2theta + theta)$ but failed to proceed . Please tell me it's approach.
trigonometry substitution
edited Aug 1 at 10:21
Michael Rozenberg
87.4k1577179
asked Aug 1 at 7:23
user580093
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My try : Sorry , but I can't figure out what is the first step to do in this question . But still I tried and converted $sin(3theta)$ into $sin (2theta + theta)$ but failed to proceed . Please tell me it's approach.
trigonometry substitution
edited Aug 1 at 10:21
Michael Rozenberg
87.4k1577179
asked Aug 1 at 7:23
user580093
204
1
Math formulas on this site would look nicer if formatted with $LaTeX$.
– user202729
Aug 1 at 7:26
1
Also, the question body should contain the question, not just the title.
– user202729
Aug 1 at 7:28
User 202729 sorry for the difficulty you faced but I don't know the latex form
– user580093
Aug 1 at 7:37
Read the link above for how to use $LaTeX$.
– user202729
Aug 1 at 7:38
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
My try : Sorry , but I can't figure out what is the first step to do in this question . But still I tried and converted $sin(3theta)$ into $sin (2theta + theta)$ but failed to proceed . Please tell me it's approach.
trigonometry substitution
edited Aug 1 at 10:21
Michael Rozenberg
87.4k1577179
asked Aug 1 at 7:23
user580093
204
My try : Sorry , but I can't figure out what is the first step to do in this question . But still I tried and converted $sin(3theta)$ into $sin (2theta + theta)$ but failed to proceed . Please tell me it's approach.
trigonometry substitution
edited Aug 1 at 10:21
Michael Rozenberg
87.4k1577179
asked Aug 1 at 7:23
user580093
204
edited Aug 1 at 10:21
Michael Rozenberg
87.4k1577179
edited Aug 1 at 10:21
Michael Rozenberg
87.4k1577179
edited Aug 1 at 10:21
Michael Rozenberg
87.4k1577179
87.4k1577179
asked Aug 1 at 7:23
user580093
204
asked Aug 1 at 7:23
user580093
204
asked Aug 1 at 7:23
user580093
204
204
1
Math formulas on this site would look nicer if formatted with $LaTeX$.
– user202729
Aug 1 at 7:26
1
Also, the question body should contain the question, not just the title.
– user202729
Aug 1 at 7:28
User 202729 sorry for the difficulty you faced but I don't know the latex form
– user580093
Aug 1 at 7:37
Read the link above for how to use $LaTeX$.
– user202729
Aug 1 at 7:38
add a comment |Â
1
Math formulas on this site would look nicer if formatted with $LaTeX$.
– user202729
Aug 1 at 7:26
1
Also, the question body should contain the question, not just the title.
– user202729
Aug 1 at 7:28
User 202729 sorry for the difficulty you faced but I don't know the latex form
– user580093
Aug 1 at 7:37
Read the link above for how to use $LaTeX$.
– user202729
Aug 1 at 7:38
1
1
Math formulas on this site would look nicer if formatted with $LaTeX$.
– user202729
Aug 1 at 7:26
Math formulas on this site would look nicer if formatted with $LaTeX$.
– user202729
Aug 1 at 7:26
1
1
Also, the question body should contain the question, not just the title.
– user202729
Aug 1 at 7:28
Also, the question body should contain the question, not just the title.
– user202729
Aug 1 at 7:28
User 202729 sorry for the difficulty you faced but I don't know the latex form
– user580093
Aug 1 at 7:37
User 202729 sorry for the difficulty you faced but I don't know the latex form
– user580093
Aug 1 at 7:37
Read the link above for how to use $LaTeX$.
– user202729
Aug 1 at 7:38
Read the link above for how to use $LaTeX$.
– user202729
Aug 1 at 7:38
add a comment |Â
5 Answers
5
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up vote
1
down vote
Let $cos2theta=x$.
Thus, we need to solve
$$3-4sin^2theta=8sin^22thetacos2theta$$ or
$$3-2(1-x)=8(1-x^2)x$$ or
$$8x^3-6x+1=0$$ or
$$2cos6theta+1=0$$ or
$$cos6theta=-frac12.$$
Can you end it now?
On $(0,pi)$ I got five roots: $$left20^circ,40^circ,100^circ,140^circ,160^circright$$
answered Aug 1 at 7:35
Michael Rozenberg
87.4k1577179
Michael but the answer is 4 real solutions
– user580093
Aug 1 at 7:38
@user580093 I fixed my post. See now.
– Michael Rozenberg
Aug 1 at 7:44
@user580093 I think your book is wrong! I got five roots. See my post.
– Michael Rozenberg
Aug 1 at 7:52
I think everyone kinda missed the sixth root $theta = 0$
– Kenny Lau
Aug 1 at 7:59
@Kenny Lau It was wrong editing. $0<theta<pi$.
– Michael Rozenberg
Aug 1 at 8:05
add a comment |Â
up vote
1
down vote
Let $u = sin theta$.
$$beginarrayrcl
sin(3theta) &=& 4 sin (theta) sin (2theta) sin (4theta) \
3u - 4u^3 &=& 4 u (2ucostheta) (4ucostheta - 8u^3costheta) \
3u - 4u^3 &=& 4 u (2u) (4u - 8u^3) (1 - u^2) \
3u - 4u^3 &=& 64 u^7 - 96 u^5 + 32 u^3 \
64 u^7 - 96 u^5 + 36 u^3 - 3u &=& 0 \
endarray$$
Since $u ne 0$, let $t = u^2$.
$$64t^3 - 96t^2 + 36t - 3 = 0$$
We depress the cubic by letting $t := x + frac12$:
$$64 x^3 - 12 x - 1 = 0$$
We consider $4(ax)^3 - 3(ax) = (4a^3) x^3 - (3a) x$ and want $a$ such that $4a^3 : -3a = 64 : -12$, i.e. $4a^2 : 3 = 16 : 3$, so we can let $a = 2$.
That means we further let $s := 2x$ and this gives us:
$$4s^3-3s = frac12$$
And then we let $s = cos(varphi)$ which gives us:
$$beginarrayrcl
cos(3varphi) &=& -dfrac12 \
cos(3arccos(s)) &=& -dfrac12 \
cos(3arccos(2x)) &=& -dfrac12 \
cos(3arccos(2t-1)) &=& -dfrac12 \
cos(3arccos(2u^2-1)) &=& -dfrac12 \
cos(3arccos(2(sintheta)^2-1)) &=& -dfrac12 \
cos(3arccos(-cos(2theta))) &=& -dfrac12 \
cos(3(pi-2theta)) &=& -dfrac12 \
cos(6theta) &=& dfrac12 \
endarray$$
answered Aug 1 at 7:50
Kenny Lau
17.7k2156
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up vote
0
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Use that
$$sin(3x)=3cos(x)^2sin(x)-sin(x)^3$$
and
$$sin(4x)=4cos(x)^3sin(x)-4cos(x)sin(x)^3$$
answered Aug 1 at 7:34
Dr. Sonnhard Graubner
66.6k32659
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Using http://mathworld.wolfram.com/WernerFormulas.html,
$$2sin t(2sin2tsin4t)=2sin t(cos2t-cos6t)$$
$$sin3t-sin t-2sin tcos6t$$
So, we have $$sin t(1+2cos6t)=0$$
answered Aug 1 at 8:32
lab bhattacharjee
214k14152263
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There are $8$ solutions, namely
$$0,quadpiover9,quad2piover9,quad4piover9,quad5piover9,quad7piover9,quad8piover9,quadpi .tag1$$
In order to arrive at them we write $e^it=:z$. The equation
$$bigl(f(t):=bigr)quad 4sin tsin(2t)sin(4t)-sin(3t)=0$$
then appears as $$z^14-z^12+z^8-z^6+z^2-1=0 ,tag2$$ or
$$(z^2-1)(z^12+z^6+1)=0 .$$
The first factor gives $zin1,-1$. For the second factor we write $z^6=:w$ and then have to solve $w^2+w+1=0$. It follows that $wine^2pi i/3, >e^-2pi i/3$. For each of these $w$-values we obtain $6$ possible values of $z$ forming a regular hexagon on the unit circle. In all we obtain $14$ different values $z_k$ satisfying $(2)$. Collecting the $z_k$ with argument $t_kinbigl[0,piover2bigr]$ leads to the list $(1)$. Here is a plot of the function $f$. In fact
$$f(t)=-sin tbigl(2cos(6t)+1bigr) .$$
answered Aug 1 at 8:39
Christian Blatter
163k7106305
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Let $cos2theta=x$.
Thus, we need to solve
$$3-4sin^2theta=8sin^22thetacos2theta$$ or
$$3-2(1-x)=8(1-x^2)x$$ or
$$8x^3-6x+1=0$$ or
$$2cos6theta+1=0$$ or
$$cos6theta=-frac12.$$
Can you end it now?
On $(0,pi)$ I got five roots: $$left20^circ,40^circ,100^circ,140^circ,160^circright$$
answered Aug 1 at 7:35
Michael Rozenberg
87.4k1577179
Michael but the answer is 4 real solutions
– user580093
Aug 1 at 7:38
@user580093 I fixed my post. See now.
– Michael Rozenberg
Aug 1 at 7:44
@user580093 I think your book is wrong! I got five roots. See my post.
– Michael Rozenberg
Aug 1 at 7:52
I think everyone kinda missed the sixth root $theta = 0$
– Kenny Lau
Aug 1 at 7:59
@Kenny Lau It was wrong editing. $0<theta<pi$.
– Michael Rozenberg
Aug 1 at 8:05
add a comment |Â
up vote
1
down vote
Let $cos2theta=x$.
Thus, we need to solve
$$3-4sin^2theta=8sin^22thetacos2theta$$ or
$$3-2(1-x)=8(1-x^2)x$$ or
$$8x^3-6x+1=0$$ or
$$2cos6theta+1=0$$ or
$$cos6theta=-frac12.$$
Can you end it now?
On $(0,pi)$ I got five roots: $$left20^circ,40^circ,100^circ,140^circ,160^circright$$
answered Aug 1 at 7:35
Michael Rozenberg
87.4k1577179
Michael but the answer is 4 real solutions
– user580093
Aug 1 at 7:38
@user580093 I fixed my post. See now.
– Michael Rozenberg
Aug 1 at 7:44
@user580093 I think your book is wrong! I got five roots. See my post.
– Michael Rozenberg
Aug 1 at 7:52
I think everyone kinda missed the sixth root $theta = 0$
– Kenny Lau
Aug 1 at 7:59
@Kenny Lau It was wrong editing. $0<theta<pi$.
– Michael Rozenberg
Aug 1 at 8:05
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $cos2theta=x$.
Thus, we need to solve
$$3-4sin^2theta=8sin^22thetacos2theta$$ or
$$3-2(1-x)=8(1-x^2)x$$ or
$$8x^3-6x+1=0$$ or
$$2cos6theta+1=0$$ or
$$cos6theta=-frac12.$$
Can you end it now?
On $(0,pi)$ I got five roots: $$left20^circ,40^circ,100^circ,140^circ,160^circright$$
answered Aug 1 at 7:35
Michael Rozenberg
87.4k1577179
Let $cos2theta=x$.
Thus, we need to solve
$$3-4sin^2theta=8sin^22thetacos2theta$$ or
$$3-2(1-x)=8(1-x^2)x$$ or
$$8x^3-6x+1=0$$ or
$$2cos6theta+1=0$$ or
$$cos6theta=-frac12.$$
Can you end it now?
On $(0,pi)$ I got five roots: $$left20^circ,40^circ,100^circ,140^circ,160^circright$$
answered Aug 1 at 7:35
Michael Rozenberg
87.4k1577179
edited Aug 1 at 8:07
answered Aug 1 at 7:35
Michael Rozenberg
87.4k1577179
answered Aug 1 at 7:35
Michael Rozenberg
87.4k1577179
answered Aug 1 at 7:35
Michael Rozenberg
87.4k1577179
87.4k1577179
Michael but the answer is 4 real solutions
– user580093
Aug 1 at 7:38
@user580093 I fixed my post. See now.
– Michael Rozenberg
Aug 1 at 7:44
@user580093 I think your book is wrong! I got five roots. See my post.
– Michael Rozenberg
Aug 1 at 7:52
I think everyone kinda missed the sixth root $theta = 0$
– Kenny Lau
Aug 1 at 7:59
@Kenny Lau It was wrong editing. $0<theta<pi$.
– Michael Rozenberg
Aug 1 at 8:05
add a comment |Â
Michael but the answer is 4 real solutions
– user580093
Aug 1 at 7:38
@user580093 I fixed my post. See now.
– Michael Rozenberg
Aug 1 at 7:44
@user580093 I think your book is wrong! I got five roots. See my post.
– Michael Rozenberg
Aug 1 at 7:52
I think everyone kinda missed the sixth root $theta = 0$
– Kenny Lau
Aug 1 at 7:59
@Kenny Lau It was wrong editing. $0<theta<pi$.
– Michael Rozenberg
Aug 1 at 8:05
Michael but the answer is 4 real solutions
– user580093
Aug 1 at 7:38
Michael but the answer is 4 real solutions
– user580093
Aug 1 at 7:38
@user580093 I fixed my post. See now.
– Michael Rozenberg
Aug 1 at 7:44
@user580093 I fixed my post. See now.
– Michael Rozenberg
Aug 1 at 7:44
@user580093 I think your book is wrong! I got five roots. See my post.
– Michael Rozenberg
Aug 1 at 7:52
@user580093 I think your book is wrong! I got five roots. See my post.
– Michael Rozenberg
Aug 1 at 7:52
I think everyone kinda missed the sixth root $theta = 0$
– Kenny Lau
Aug 1 at 7:59
I think everyone kinda missed the sixth root $theta = 0$
– Kenny Lau
Aug 1 at 7:59
@Kenny Lau It was wrong editing. $0<theta<pi$.
– Michael Rozenberg
Aug 1 at 8:05
@Kenny Lau It was wrong editing. $0<theta<pi$.
– Michael Rozenberg
Aug 1 at 8:05
add a comment |Â
up vote
1
down vote
Let $u = sin theta$.
$$beginarrayrcl
sin(3theta) &=& 4 sin (theta) sin (2theta) sin (4theta) \
3u - 4u^3 &=& 4 u (2ucostheta) (4ucostheta - 8u^3costheta) \
3u - 4u^3 &=& 4 u (2u) (4u - 8u^3) (1 - u^2) \
3u - 4u^3 &=& 64 u^7 - 96 u^5 + 32 u^3 \
64 u^7 - 96 u^5 + 36 u^3 - 3u &=& 0 \
endarray$$
Since $u ne 0$, let $t = u^2$.
$$64t^3 - 96t^2 + 36t - 3 = 0$$
We depress the cubic by letting $t := x + frac12$:
$$64 x^3 - 12 x - 1 = 0$$
We consider $4(ax)^3 - 3(ax) = (4a^3) x^3 - (3a) x$ and want $a$ such that $4a^3 : -3a = 64 : -12$, i.e. $4a^2 : 3 = 16 : 3$, so we can let $a = 2$.
That means we further let $s := 2x$ and this gives us:
$$4s^3-3s = frac12$$
And then we let $s = cos(varphi)$ which gives us:
$$beginarrayrcl
cos(3varphi) &=& -dfrac12 \
cos(3arccos(s)) &=& -dfrac12 \
cos(3arccos(2x)) &=& -dfrac12 \
cos(3arccos(2t-1)) &=& -dfrac12 \
cos(3arccos(2u^2-1)) &=& -dfrac12 \
cos(3arccos(2(sintheta)^2-1)) &=& -dfrac12 \
cos(3arccos(-cos(2theta))) &=& -dfrac12 \
cos(3(pi-2theta)) &=& -dfrac12 \
cos(6theta) &=& dfrac12 \
endarray$$
answered Aug 1 at 7:50
Kenny Lau
17.7k2156
add a comment |Â
up vote
1
down vote
Let $u = sin theta$.
$$beginarrayrcl
sin(3theta) &=& 4 sin (theta) sin (2theta) sin (4theta) \
3u - 4u^3 &=& 4 u (2ucostheta) (4ucostheta - 8u^3costheta) \
3u - 4u^3 &=& 4 u (2u) (4u - 8u^3) (1 - u^2) \
3u - 4u^3 &=& 64 u^7 - 96 u^5 + 32 u^3 \
64 u^7 - 96 u^5 + 36 u^3 - 3u &=& 0 \
endarray$$
Since $u ne 0$, let $t = u^2$.
$$64t^3 - 96t^2 + 36t - 3 = 0$$
We depress the cubic by letting $t := x + frac12$:
$$64 x^3 - 12 x - 1 = 0$$
We consider $4(ax)^3 - 3(ax) = (4a^3) x^3 - (3a) x$ and want $a$ such that $4a^3 : -3a = 64 : -12$, i.e. $4a^2 : 3 = 16 : 3$, so we can let $a = 2$.
That means we further let $s := 2x$ and this gives us:
$$4s^3-3s = frac12$$
And then we let $s = cos(varphi)$ which gives us:
$$beginarrayrcl
cos(3varphi) &=& -dfrac12 \
cos(3arccos(s)) &=& -dfrac12 \
cos(3arccos(2x)) &=& -dfrac12 \
cos(3arccos(2t-1)) &=& -dfrac12 \
cos(3arccos(2u^2-1)) &=& -dfrac12 \
cos(3arccos(2(sintheta)^2-1)) &=& -dfrac12 \
cos(3arccos(-cos(2theta))) &=& -dfrac12 \
cos(3(pi-2theta)) &=& -dfrac12 \
cos(6theta) &=& dfrac12 \
endarray$$
answered Aug 1 at 7:50
Kenny Lau
17.7k2156
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $u = sin theta$.
$$beginarrayrcl
sin(3theta) &=& 4 sin (theta) sin (2theta) sin (4theta) \
3u - 4u^3 &=& 4 u (2ucostheta) (4ucostheta - 8u^3costheta) \
3u - 4u^3 &=& 4 u (2u) (4u - 8u^3) (1 - u^2) \
3u - 4u^3 &=& 64 u^7 - 96 u^5 + 32 u^3 \
64 u^7 - 96 u^5 + 36 u^3 - 3u &=& 0 \
endarray$$
Since $u ne 0$, let $t = u^2$.
$$64t^3 - 96t^2 + 36t - 3 = 0$$
We depress the cubic by letting $t := x + frac12$:
$$64 x^3 - 12 x - 1 = 0$$
We consider $4(ax)^3 - 3(ax) = (4a^3) x^3 - (3a) x$ and want $a$ such that $4a^3 : -3a = 64 : -12$, i.e. $4a^2 : 3 = 16 : 3$, so we can let $a = 2$.
That means we further let $s := 2x$ and this gives us:
$$4s^3-3s = frac12$$
And then we let $s = cos(varphi)$ which gives us:
$$beginarrayrcl
cos(3varphi) &=& -dfrac12 \
cos(3arccos(s)) &=& -dfrac12 \
cos(3arccos(2x)) &=& -dfrac12 \
cos(3arccos(2t-1)) &=& -dfrac12 \
cos(3arccos(2u^2-1)) &=& -dfrac12 \
cos(3arccos(2(sintheta)^2-1)) &=& -dfrac12 \
cos(3arccos(-cos(2theta))) &=& -dfrac12 \
cos(3(pi-2theta)) &=& -dfrac12 \
cos(6theta) &=& dfrac12 \
endarray$$
answered Aug 1 at 7:50
Kenny Lau
17.7k2156
Let $u = sin theta$.
$$beginarrayrcl
sin(3theta) &=& 4 sin (theta) sin (2theta) sin (4theta) \
3u - 4u^3 &=& 4 u (2ucostheta) (4ucostheta - 8u^3costheta) \
3u - 4u^3 &=& 4 u (2u) (4u - 8u^3) (1 - u^2) \
3u - 4u^3 &=& 64 u^7 - 96 u^5 + 32 u^3 \
64 u^7 - 96 u^5 + 36 u^3 - 3u &=& 0 \
endarray$$
Since $u ne 0$, let $t = u^2$.
$$64t^3 - 96t^2 + 36t - 3 = 0$$
We depress the cubic by letting $t := x + frac12$:
$$64 x^3 - 12 x - 1 = 0$$
We consider $4(ax)^3 - 3(ax) = (4a^3) x^3 - (3a) x$ and want $a$ such that $4a^3 : -3a = 64 : -12$, i.e. $4a^2 : 3 = 16 : 3$, so we can let $a = 2$.
That means we further let $s := 2x$ and this gives us:
$$4s^3-3s = frac12$$
And then we let $s = cos(varphi)$ which gives us:
$$beginarrayrcl
cos(3varphi) &=& -dfrac12 \
cos(3arccos(s)) &=& -dfrac12 \
cos(3arccos(2x)) &=& -dfrac12 \
cos(3arccos(2t-1)) &=& -dfrac12 \
cos(3arccos(2u^2-1)) &=& -dfrac12 \
cos(3arccos(2(sintheta)^2-1)) &=& -dfrac12 \
cos(3arccos(-cos(2theta))) &=& -dfrac12 \
cos(3(pi-2theta)) &=& -dfrac12 \
cos(6theta) &=& dfrac12 \
endarray$$
answered Aug 1 at 7:50
Kenny Lau
17.7k2156
edited Aug 1 at 8:11
answered Aug 1 at 7:50
Kenny Lau
17.7k2156
answered Aug 1 at 7:50
Kenny Lau
17.7k2156
answered Aug 1 at 7:50
Kenny Lau
17.7k2156
17.7k2156
add a comment |Â
add a comment |Â
up vote
0
down vote
Use that
$$sin(3x)=3cos(x)^2sin(x)-sin(x)^3$$
and
$$sin(4x)=4cos(x)^3sin(x)-4cos(x)sin(x)^3$$
answered Aug 1 at 7:34
Dr. Sonnhard Graubner
66.6k32659
add a comment |Â
up vote
0
down vote
Use that
$$sin(3x)=3cos(x)^2sin(x)-sin(x)^3$$
and
$$sin(4x)=4cos(x)^3sin(x)-4cos(x)sin(x)^3$$
answered Aug 1 at 7:34
Dr. Sonnhard Graubner
66.6k32659
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Use that
$$sin(3x)=3cos(x)^2sin(x)-sin(x)^3$$
and
$$sin(4x)=4cos(x)^3sin(x)-4cos(x)sin(x)^3$$
answered Aug 1 at 7:34
Dr. Sonnhard Graubner
66.6k32659
Use that
$$sin(3x)=3cos(x)^2sin(x)-sin(x)^3$$
and
$$sin(4x)=4cos(x)^3sin(x)-4cos(x)sin(x)^3$$
answered Aug 1 at 7:34
Dr. Sonnhard Graubner
66.6k32659
answered Aug 1 at 7:34
Dr. Sonnhard Graubner
66.6k32659
answered Aug 1 at 7:34
Dr. Sonnhard Graubner
66.6k32659
answered Aug 1 at 7:34
Dr. Sonnhard Graubner
66.6k32659
66.6k32659
add a comment |Â
add a comment |Â
up vote
0
down vote
Using http://mathworld.wolfram.com/WernerFormulas.html,
$$2sin t(2sin2tsin4t)=2sin t(cos2t-cos6t)$$
$$sin3t-sin t-2sin tcos6t$$
So, we have $$sin t(1+2cos6t)=0$$
answered Aug 1 at 8:32
lab bhattacharjee
214k14152263
add a comment |Â
up vote
0
down vote
Using http://mathworld.wolfram.com/WernerFormulas.html,
$$2sin t(2sin2tsin4t)=2sin t(cos2t-cos6t)$$
$$sin3t-sin t-2sin tcos6t$$
So, we have $$sin t(1+2cos6t)=0$$
answered Aug 1 at 8:32
lab bhattacharjee
214k14152263
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Using http://mathworld.wolfram.com/WernerFormulas.html,
$$2sin t(2sin2tsin4t)=2sin t(cos2t-cos6t)$$
$$sin3t-sin t-2sin tcos6t$$
So, we have $$sin t(1+2cos6t)=0$$
answered Aug 1 at 8:32
lab bhattacharjee
214k14152263
Using http://mathworld.wolfram.com/WernerFormulas.html,
$$2sin t(2sin2tsin4t)=2sin t(cos2t-cos6t)$$
$$sin3t-sin t-2sin tcos6t$$
So, we have $$sin t(1+2cos6t)=0$$
answered Aug 1 at 8:32
lab bhattacharjee
214k14152263
answered Aug 1 at 8:32
lab bhattacharjee
214k14152263
answered Aug 1 at 8:32
lab bhattacharjee
214k14152263
answered Aug 1 at 8:32
lab bhattacharjee
214k14152263
214k14152263
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add a comment |Â
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There are $8$ solutions, namely
$$0,quadpiover9,quad2piover9,quad4piover9,quad5piover9,quad7piover9,quad8piover9,quadpi .tag1$$
In order to arrive at them we write $e^it=:z$. The equation
$$bigl(f(t):=bigr)quad 4sin tsin(2t)sin(4t)-sin(3t)=0$$
then appears as $$z^14-z^12+z^8-z^6+z^2-1=0 ,tag2$$ or
$$(z^2-1)(z^12+z^6+1)=0 .$$
The first factor gives $zin1,-1$. For the second factor we write $z^6=:w$ and then have to solve $w^2+w+1=0$. It follows that $wine^2pi i/3, >e^-2pi i/3$. For each of these $w$-values we obtain $6$ possible values of $z$ forming a regular hexagon on the unit circle. In all we obtain $14$ different values $z_k$ satisfying $(2)$. Collecting the $z_k$ with argument $t_kinbigl[0,piover2bigr]$ leads to the list $(1)$. Here is a plot of the function $f$. In fact
$$f(t)=-sin tbigl(2cos(6t)+1bigr) .$$
answered Aug 1 at 8:39
Christian Blatter
163k7106305
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up vote
0
down vote
There are $8$ solutions, namely
$$0,quadpiover9,quad2piover9,quad4piover9,quad5piover9,quad7piover9,quad8piover9,quadpi .tag1$$
In order to arrive at them we write $e^it=:z$. The equation
$$bigl(f(t):=bigr)quad 4sin tsin(2t)sin(4t)-sin(3t)=0$$
then appears as $$z^14-z^12+z^8-z^6+z^2-1=0 ,tag2$$ or
$$(z^2-1)(z^12+z^6+1)=0 .$$
The first factor gives $zin1,-1$. For the second factor we write $z^6=:w$ and then have to solve $w^2+w+1=0$. It follows that $wine^2pi i/3, >e^-2pi i/3$. For each of these $w$-values we obtain $6$ possible values of $z$ forming a regular hexagon on the unit circle. In all we obtain $14$ different values $z_k$ satisfying $(2)$. Collecting the $z_k$ with argument $t_kinbigl[0,piover2bigr]$ leads to the list $(1)$. Here is a plot of the function $f$. In fact
$$f(t)=-sin tbigl(2cos(6t)+1bigr) .$$
answered Aug 1 at 8:39
Christian Blatter
163k7106305
add a comment |Â
up vote
0
down vote
up vote
0
down vote
There are $8$ solutions, namely
$$0,quadpiover9,quad2piover9,quad4piover9,quad5piover9,quad7piover9,quad8piover9,quadpi .tag1$$
In order to arrive at them we write $e^it=:z$. The equation
$$bigl(f(t):=bigr)quad 4sin tsin(2t)sin(4t)-sin(3t)=0$$
then appears as $$z^14-z^12+z^8-z^6+z^2-1=0 ,tag2$$ or
$$(z^2-1)(z^12+z^6+1)=0 .$$
The first factor gives $zin1,-1$. For the second factor we write $z^6=:w$ and then have to solve $w^2+w+1=0$. It follows that $wine^2pi i/3, >e^-2pi i/3$. For each of these $w$-values we obtain $6$ possible values of $z$ forming a regular hexagon on the unit circle. In all we obtain $14$ different values $z_k$ satisfying $(2)$. Collecting the $z_k$ with argument $t_kinbigl[0,piover2bigr]$ leads to the list $(1)$. Here is a plot of the function $f$. In fact
$$f(t)=-sin tbigl(2cos(6t)+1bigr) .$$
answered Aug 1 at 8:39
Christian Blatter
163k7106305
There are $8$ solutions, namely
$$0,quadpiover9,quad2piover9,quad4piover9,quad5piover9,quad7piover9,quad8piover9,quadpi .tag1$$
In order to arrive at them we write $e^it=:z$. The equation
$$bigl(f(t):=bigr)quad 4sin tsin(2t)sin(4t)-sin(3t)=0$$
then appears as $$z^14-z^12+z^8-z^6+z^2-1=0 ,tag2$$ or
$$(z^2-1)(z^12+z^6+1)=0 .$$
The first factor gives $zin1,-1$. For the second factor we write $z^6=:w$ and then have to solve $w^2+w+1=0$. It follows that $wine^2pi i/3, >e^-2pi i/3$. For each of these $w$-values we obtain $6$ possible values of $z$ forming a regular hexagon on the unit circle. In all we obtain $14$ different values $z_k$ satisfying $(2)$. Collecting the $z_k$ with argument $t_kinbigl[0,piover2bigr]$ leads to the list $(1)$. Here is a plot of the function $f$. In fact
$$f(t)=-sin tbigl(2cos(6t)+1bigr) .$$
answered Aug 1 at 8:39
Christian Blatter
163k7106305
edited Aug 1 at 9:34
answered Aug 1 at 8:39
Christian Blatter
163k7106305
answered Aug 1 at 8:39
Christian Blatter
163k7106305
answered Aug 1 at 8:39
Christian Blatter
163k7106305
163k7106305
add a comment |Â
add a comment |Â
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1
Math formulas on this site would look nicer if formatted with $LaTeX$.
– user202729
Aug 1 at 7:26
1
Also, the question body should contain the question, not just the title.
– user202729
Aug 1 at 7:28
User 202729 sorry for the difficulty you faced but I don't know the latex form
– user580093
Aug 1 at 7:37
Read the link above for how to use $LaTeX$.
– user202729
Aug 1 at 7:38