help with solving sin4x = -2sin2x [0, 2pi)

[tryinghard]

I need some major help solving sin4x = -2sin2x [0, 2pi)

 

[stapel]

I need some major help solving sin4x = -2sin2x [0, 2pi)

Your formatting is ambiguous. Do you mean any of the following?

. . . . .sin⁴(x) = -2 sin²(x)

. . . . .sin⁴(x) = -2 sin(2x)

. . . . .sin(4x) = -2 sin²(x)

. . . . .sin(4x) = -2 sin(2x)

Or did you mean something else?

When you reply, please show the steps you have tried so far. Thank you.

Eliz.

 

[galactus]

If you mean $\displaystyle sin(4x)=-2sin(2x)$, then divide by sin(2x).

$\displaystyle \frac{sin(4x)}{sin(2x)}=2cos(2x)$

You have: $\displaystyle 2cos(2x)=-2$

$\displaystyle cos(2x)=-1$