Math Help - trigonometry

1. trigonometry

sin6x+3sin4x.cos2x=0
find solution set

sin6x+3/2(sin6x+sin2x)=0
5sin6x+3sin2x=0

I got stuck here.

2. Re: trigonometry

Originally Posted by kastamonu
sin6x+3sin4x.cos2x=0
find solution set

sin6x+3/2(sin6x+sin2x)=0
5sin6x+3sin2x=0

I got stuck here.
Is it $sin(2x)$ or $sin^2(x)$?

-Dan

3. Re: trigonometry

write

sin 6x = sin( 4x + 2x) = ......

sin2x

5. Re: trigonometry

Originally Posted by Idea
write

sin 6x = sin( 4x + 2x) = ......
I couldn't get anything.

6. Re: trigonometry

$\sin (4x+2x)=\sin 4x \cos 2x +\cos 4x \sin 2x$

7. Re: trigonometry

Let y = 2x so we have 5.sin(3y) + 3.sin(y) = 0

where sin(3y) = sin(2y+y) = sin(2y).cos(y) + cos(2y).sin(y)

= 2.sin(y).cos^2(y) + sin(y).(1-2.sin^2(y))

= 2.sin(y)(1-sin^2(y)) + sin(y) - 2.sin^3(y)

= 3.sin(y) - 4.sin^3(y)

8. Re: trigonometry

Plug that in for sin(3y). Then, let $u = \sin(y)$. You wind up with a cubic equation in $u$. Solve for $u$ (you have something for the form $au-bu^3=0$, so $u(a-bu^2)=0$ implies $u=0$ or $a-bu^2=0$). Once you solve for $u$, you have $u = \sin(y)$, so take arcsin of both sides. Then, you have $y = 2x$.

9. Re: trigonometry

you could also linearize the equation

$\sin 2x (4+ 5\cos 4x)=0$

giving two sets of solutions

$\sin 2x=0$

$\cos 4x=-4/5$

Thanks.