What is the general solution of sin^2 x+ sin^2 2x=sin^2 3x?

Answer: \(x = \dfrac{n\pi}{2}\) or \(x = n\pi \pm \dfrac{\pi}{6}\), where \(n \in \mathbb{Z}\)

Use the identity \(\sin^{2}A - \sin^{2}B = \sin(A+B)\sin(A-B)\). Rewrite the equation as

\[\sin^{2}3x - \sin^{2}x = \sin^{2}2x\]

\[\sin 4x\,\sin 2x = \sin^{2}2x\]

\[\sin 2x\,(\sin 4x - \sin 2x) = 0\]

Since \(\sin 4x = 2\sin 2x\cos 2x\):

\[\sin^{2}2x\,(2\cos 2x - 1) = 0\]

Case 1: \(\sin 2x = 0 \Rightarrow 2x = n\pi \Rightarrow x = \dfrac{n\pi}{2}\).

Case 2: \(\cos 2x = \dfrac{1}{2} = \cos\dfrac{\pi}{3}\)

\[2x = 2n\pi \pm \frac{\pi}{3} \Rightarrow x = n\pi \pm \frac{\pi}{6}\]

Check: \(x = \pi/6\) gives \(\tfrac14 + \tfrac34 = 1 = \sin^{2}\tfrac{\pi}{2}\), which is true.