Question

Solve: ${\sin }^2\theta -\cos \theta =\frac{1}{4}$ in the interval $0\le \theta \le 2\pi$.

Answer 1

As given the equation reduces to $4{\cos }^2\theta +4cos\theta -3=0$

or $(2\cos \theta +3)(2\cos \theta -1)=0$

But $\cos \theta =-3/2$(rejected)

$\therefore \cos \theta =\frac{1}{2}=\cos \frac{\pi }{3}$

$\therefore \theta =2n\pi \pm \pi /3$

We have to choose values of $\theta$ s.t. $0\le \theta \le 2\pi$

$\therefore \theta =\pi /3,2\pi -\pi /3=5\pi /3$.