Question

If $2{\sin }^2\theta =3\cos \theta$, where $0\le \theta \le 2\pi ,$ then $\theta =$

• A. $\frac{\pi }{6},\frac{7\pi }{6}$
• B. $\frac{\pi }{3},\frac{5\pi }{3}$
• C. $\frac{\pi }{3},\frac{7\pi }{3}$
• D. None of these

Answer 1

Answer: (B)

$\frac{\pi }{3},\frac{5\pi }{3}$

Given: $2{\sin }^2\theta =3\cos \theta$

$\Rightarrow 2(1-{\cos }^2\theta )=3\cos \theta$

$\Rightarrow 2{\cos }^2\theta +3\cos \theta -2=0$

$\Rightarrow 2{\cos }^2\theta +4\cos \theta -\cos \theta -2=0$

$\Rightarrow (2\cos \theta -1)(\cos \theta +2)=0$

$\Rightarrow \cos \theta =\frac{1}{2}$ and $\cos \theta =-2$ is not possible because $\cos \theta \ge -1$

$\Rightarrow \cos \theta =\frac{1}{2}$

$\Rightarrow \theta =\frac{\pi }{3},\frac{5\pi }{3}$