# Solve Sin6x+sin4x

## Homework Statement

$$sin6x+sin4x=0$$

## Homework Equations

$$sinx=2sin\frac{x}{2}cos\frac{x}{2}$$

$$sin2x=2sinxcosx$$

$$cos2x=cos^2x-sin^x$$

## The Attempt at a Solution

$$2sin3xcos3x+2sin2xcos2x=0$$

$$sin3xcos3x+sin2xcos2x=0$$

What shall I do next?

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Defennder
Homework Helper
What exactly are you suppose to do? What is the question?

EDIT: Is it to solve for x?

Yes. I need to find x.

Ohh... Can I solve it like this:

$$sin6x=-sin4x$$

$$sin6x=sin(-4x)$$

$$6x=-4x+2k\pi$$

$$10x=2k\pi$$

$$x=\frac{k\pi}{5}$$

??

Defennder
Homework Helper
That is partially correct. But you're missing out on other possible values of x. $$x=\frac{\pi}{2}$$ also satisfies the equation but it's not expressible in your answer.

Use this trigo identity:
$$2sin(Ax)cos(Bx) = sin((A-B)x) + sin((A+B)x)$$

Yes I forgot.

$$6x=\pi+4x+2k\pi$$

$$x=\frac{\pi}{2}+k\pi$$

Defennder
Homework Helper
Where did pi in your first equation come from?

Remember this:

x=arcsinx+2kpi

x=pi - arcsinx + 2kpi

?