Question

If ${\sin }^2\theta -2\cos \theta +\frac{1}{4}=0,$ then the general value of $\theta$ is :

Answer 1

$\begin{array}{c}{\sin }^2\theta -2\cos \theta +\frac{1}{4}=0\\ 4-4{\cos }^2\theta -8\cos \theta +1=0\\ 4{\cos }^2\theta +8\cos \theta -5=0\\ \left(4{\cos }^2\theta +10\cos \theta \right)-\left(2\cos \theta +5\right)=0\\ 2\cos \theta \left(2\cos \theta +5\right)-\left(2\cos \theta +5\right)=0\\ \left(2\cos \theta -1\right)\left(2\cos \theta +5\right)=0\\ 2\cos \theta =1\\ \cos \theta =\frac{1}{2}\\ =\cos =\frac{1}{2}\\ =\cos \frac{\pi }{3}\\ \therefore Q=2n\pi \pm \frac{\pi }{3}\end{array}$