Verified answer

Remember that sin(2θ) = 2sin(θ)cos(θ), so this becomes

2sin(θ)cos(θ) = 2sin^2(θ)

Before we divide both sides by sin(θ), we have to remember that having sin(θ) = 0 gives us solutions. So there's one set of solutions so far: θ = 0, ±π, ±2π, ±3π, etc.

Now assuming that sin(θ) is not zero, we can divide both sides by it to get:

2cos(θ) = 2sin(θ)

cos(θ) = sin(θ)

tan(θ) = 1

This holds for θ = π/4, 5π/4, etc.

Ummm, where is the x?

What you are probably after are double angle formulas (part of trig identities). Check out Wikipedia (see below)

In particular you probably want...

sin2θ= 2sinθcosθ

So to solve your equation it comes down to...

sinθ = cosθ

which is true for θ = 45 + 180n (n integer)

Sin2θ = 2sin²θ OR sinθ{cosθ -- sinθ) = 0 whence θ = 0, 45 degrees.