Question

Solutions of the equation ${\sin }^2\theta -\cos \theta =\frac{1}{4},0\le \theta \le 2\pi$

• A. $\frac{2\pi }{3},\frac{\pi }{3}$
• B. $\frac{\pi }{3},\frac{5\pi }{3}$
• C. $-\frac{\pi }{3},\frac{2\pi }{3}$
• D. $\frac{2\pi }{3},\frac{5\pi }{3}$

Answer 1

Answer: (B)

${\sin }^2\theta -\cos \theta =\frac{1}{4}$
$\Rightarrow 4(1-{\cos }^2\theta )-4\cos \theta =1$
$\Rightarrow 4{\cos }^2\theta +4\cos \theta -3=0$
$\Rightarrow (2cos\theta +3)(2\cos \theta -1)=0$
$\Rightarrow \cos \theta =-\frac{3}{2}$ or $\cos \theta =\frac{1}{2}$
But $\cos \theta =-\frac{3}{2}$ is not possible
$\therefore \cos \theta =\frac{1}{2}=\cos \frac{\pi }{3}$ or $\cos \left(2\pi -\frac{\pi }{3}\right)$
$=\cos \frac{\pi }{3}$ or $\cos \frac{5\pi }{3}(0\le \theta \le 2\pi )$
$\therefore \theta =\frac{\pi }{3},\frac{5\pi }{3}$