Express $3\sin2x-4\cos2x$ as $R\sin(2x-\alpha)$, where $\alpha$ is acute. Hence solve \[ 3\sin

Question

Express $3\sin2x-4\cos2x$ as $R\sin(2x-\alpha)$, where $\alpha$ is acute. Hence solve $$ 3\sin2x-4\cos2x=0,\qquad 0\le x\le\pi. $$

Accepted Answer

(a) $5\sin(2x-0.927)$; (b) $x=0.464,2.03$ rad.

Trigonometry