Mathematics - Trigonometric Equation Question with Solution | TestHub

$$\begin{array}{llccc}\therefore & 2{\sin }^2\theta -\cos 2\theta =0\Rightarrow 1-\cos 2\theta =0\Rightarrow \cos 2\theta =\frac{1}{2}& 2\theta =\frac{\pi }{3},\frac{5\pi }{3},\frac{7\pi }{3},\frac{11\pi }{3}\Rightarrow \theta =\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6}\therefore & 2{\cos }^2\theta -3\sin \theta =0\therefore & (2\sin \theta -1)(\sin \theta +2)=0\Rightarrow \sin \theta =\frac{1}{2}\end{array}$$

$\therefore \theta =\frac{\pi }{6},\frac{5\pi }{6}\Rightarrow$ No.of solutions $=2$