Question

sin2x=sinx, 0<=x<=2pi

Answer 1

Hint: Subtract sin x from both sides to get

\(\sin^2x - \sin x = 0\)

Then factor out sin x common to both terms to get

\(\sin x (\sin x - 1) =0\)

Now use zero product property again.

Answer 2

i think its sin2x not \[\sin ^{2}x\]

Answer 3

Okay, so if it's \(\sin(2x)\), then that is equal to \(2\sin(x)\cos(x)\)

Answer 4

Which means

\(\sin(2x) = \sin x\)

Subtract sin x from both sides to get

\(\sin(2x) - \sin x = 0\)

Replace sin(2x) with 2sin x cos x to get

\(2\sin x \cos x - \sin x = 0\)

Now factor out sin x common to both terms:

\(\sin x(2\cos x-1) = 0\)

Now use zero product property