Question

The number of solution of the pairs of equations $2{\sin }^2\theta -\cos 2\theta =0,2{\cos }^2\theta -3\sin \theta =0$ in the interval $[0,2\pi ]$ is

• A. zero
• B. one
• C. two
• D. four

Answer 1

Answer: (C)

two

.Given are a pair of equation

$2\sin {}^2\theta -\cos 2\theta =02\cos {}^2\theta -3\sin \theta =0$

And $\theta ϵ[0,2\pi ]$

Now,

$2\sin {}^2\theta -\cos 2\theta =0\Rightarrow 2\sin {}^2\theta -1+2\sin {}^2\theta =0\Rightarrow 4\sin {}^2\theta =1\sin \theta =\pm \frac{1}{2}$

$2\cos {}^2\theta -3\sin \theta =0\Rightarrow 2-2\sin {}^2\theta -3\sin \theta =0\Rightarrow 2\sin {}^2\theta +3\sin \theta -2=0\Rightarrow (\sin \theta +2)(2\sin \theta -1)=0\Rightarrow \sin \theta =\frac{1}{2}$

Hence $\theta$ could be $\frac{\pi }{6},\frac{5\pi }{6}$.