Question

If the sum of solutions of the system of equations $2{\sin }^2\theta -\cos 2\theta =0$ and $2{\cos }^2\theta +3\sin \theta =0$ in the interval $[0,2\pi ]$ is $k\pi$, then $k$ is equal to _____

Answer 1

Answer: 3

$2{\sin }^2\theta -\cos 2\theta =0$
$2{\sin }^2\theta -\left(1-2{\sin }^2\theta \right)=0$
$\Rightarrow {\sin }^2\theta ={\left(\frac{1}{2}\right)}^2$
$\theta =\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6}$
$2{\cos }^2\theta +3\sin \theta =0$
$\Rightarrow 2{\sin }^2\theta -3\sin \theta -2=0$
$\therefore \sin \theta =-\frac{1}{2}$
$\theta =\frac{7\pi }{6},\frac{11\pi }{6}$
So, the common solution is
$\theta =\frac{7\pi }{6},\frac{11\pi }{6}$
Sum $=\frac{7\pi +11\pi }{6}=3\pi =k\pi$
$K=3$