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 $\theta \in [0,2\pi ]\}$