Question

The number of solutions of the pair 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

• A. zero
• B. one
• C. two
• D. four

Answer 1

Answer: (C)

Since we have
$2 \cos ^{2} \theta-3 \sin \theta=0$ (1)
we can write as
$2 \sin ^{2} \theta-\left(1-2 \sin ^{2} \theta\right)=0$
$4 \sin ^{2} \theta=1$
$\sin \theta=\frac{1}{2},-\frac{1}{2}$
Now, $2 \cos ^{2} \theta-3 \sin \theta=0$ (2)
$2-2 \sin ^{2} \theta-3 \sin \theta=0$
$2 \sin ^{2} \theta+3 \sin \theta-2=0$
$\sin x=\frac{1}{2},-2$
$\Rightarrow \sin \theta=\frac{1}{2}$ is common solution.
Therefore, the number of solutions is two:
$\theta=\frac{\pi}{6} $ and $ \frac{5 \pi}{6}\{$ where $ \theta \in[0,2 \pi]\}$