# Thread: Help me to solve the trigonometric equation

1. ## Help me to solve the trigonometric equation

Hi I have tried to solve this... (but my teacher never gets satisfied).

Solve the trigonometric equation:
4sinx*sin2x*sin4x=sin3x

My solution so far...

4sinx*sin2x*sin4x−sin3x=0

3x=sinx(3-4((sinx)^2)
0=4sinx*sin2x*sin4x-sinx(3-4*((sinx)^2)= sinx(4sin(2x)) *sin4x-(3-4*((sinx)^2)

So:
0=sinx

or
0=4sin2x*sin4x-(3-4*((sinx)^2)

2. Originally Posted by sf1903
Solve the trigonometric equation:
4sinx*sin2x*sin4x=sin3x
Can I suggest

$\displaystyle \sin 2x = 2\sin x \cos x$

$\displaystyle \sin 3x = 3\sin x -4 \sin^3 x$

$\displaystyle \sin 4x = 2\sin 2x \cos 2x = 2(2\sin x \cos x )(1-2\sin^2 x)$

3. Originally Posted by pickslides
Can I suggest

$\displaystyle \sin 2x = 2\sin x \cos x$

$\displaystyle \sin 3x = 3\sin x -4 \sin^3 x$

$\displaystyle \sin 4x = 2\sin 2x \cos 2x = 2(2\sin x \cos x )(1-2\sin^2 x)$

Should I then insert it to the original equation? 4sinx*sin2x*sin4x=sin3x

So:
$4*Sinx(2SinxCosx)*2(2SinxCosx)(1-2Sin^2x)=3Sinx-4Sin^3x$

$4Sinx*3(2Sinx*Cosx)(1-2*Sin^2x)=3Sinx-4Sin^3x$

$4Sinx*3(2SinxCosx)(1-2Sin^2x)-3Sinx+4Sin^3x=0$

?

4. Originally Posted by sf1903
Hi I have tried to solve this... (but my teacher never gets satisfied).

Solve the trigonometric equation:
4sinx*sin2x*sin4x=sin3x

My solution so far...

$4\sin x\sin2x\sin4x-\sin3x=0$

$\sin3x=\sin x(3-4\sin^2x)$

$0=4\sin x\sin2x\sin4x-\sin x(3-4\sin^2x)= \sin x(4\sin2x\sin4x-(3-4\sin^2x))$

So:

$0=\sin x$
or
$0=4\sin2x\sin4x-(3-4\sin^2x)$
Good start! The first alternative, $0=\sin x$, is easy to deal with, so let's look at the second one, $0 = 4\sin2x\sin4x - (3-4\sin^2x)$.

Use the trig identity $\cos(\theta-\phi) + \cos(\theta+\phi) = 2\sin\theta\sin\phi$. It tells you that $2\sin2x\sin4x = \cos2x - \cos6x$, and $2\sin^2x = 1-\cos2x$. Therefore

$0 = 4\sin2x\sin4x - (3-4\sin^2x) = 2(\cos2x - \cos6x) - (3 - 2(1-\cos2x)) = -2\cos6x-1.$

Now finish it off from there.

5. Originally Posted by sf1903
Should I then insert it to the original equation? 4sinx*sin2x*sin4x=sin3x

So:
$4*Sinx(2SinxCosx)*2(2SinxCosx)(1-2Sin^2x)=3Sinx-4Sin^3x$

$4Sinx*3(2Sinx*Cosx)(1-2*Sin^2x)=3Sinx-4Sin^3x$

$4Sinx*3(2SinxCosx)(1-2Sin^2x)-3Sinx+4Sin^3x=0$

?

This expands further...

6. Originally Posted by Opalg
Good start! The first alternative, $0=\sin x$, is easy to deal with, so let's look at the second one, $0 = 4\sin2x\sin4x - (3-4\sin^2x)$.

Use the trig identity $\cos(\theta-\phi) + \cos(\theta+\phi) = 2\sin\theta\sin\phi$. It tells you that $2\sin2x\sin4x = \cos2x - \cos6x$, and $2\sin^2x = 1-\cos2x$. Therefore

$0 = 4\sin2x\sin4x - (3-4\sin^2x) = 2(\cos2x - \cos6x) - (3 - 2(1-\cos2x) = -2\cos6x-1.$

Now finish it off from there.
So:
$2(\cos2x - \cos6x) - (3 - 2(1-\cos2x) = -2\cos6x-1.$

Therefore:
$2(\cos2x - \cos6x) - (3 - 2(1-\cos2x)+2\cos6x-1=0
$

How should I continue?

7. Originally Posted by sf1903
So:
$2(\cos2x - \cos6x) - (3 - 2(1-\cos2x) = -2\cos6x-1.$

Therefore:
$2(\cos2x - \cos6x) - (3 - 2(1-\cos2x)+2\cos6x-1=0
$

How should I continue?
What's with the ? You have been told exactly how to do it!:

Originally Posted by opalg
$0 = 4\sin2x\sin4x - (3-4\sin^2x) = 2(\cos2x - \cos6x) - (3 - 2(1-\cos2x) = -2\cos6x-1.$
So the equation you have to solve is obviously $0 = -2 \cos(6x) - 1$. Surely you can solve this ....

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