Question

Solve the equations
${\sin }^2\theta -2\cos \theta +\frac{1}{4}=0$.

Answer 1

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

$\Rightarrow 4(1-{\cos }^2\theta )-2\cos \theta +1=0$

$\Rightarrow 4-4{\cos }^2\theta -8\cos \theta +1=0$

$\Rightarrow -4{\cos }^2\theta -8\cos \theta +5=0$

$\Rightarrow 4{\cos }^2\theta +8\cos \theta -5=0$

$\Rightarrow 4{\cos }^2\theta +10\cos \theta -2\cos \theta -5=0$

$\Rightarrow 2\cos \theta (2\cos \theta +5)-1(2\cos \theta +5)=0$

$\Rightarrow (2\cos \theta +5)(2\cos \theta -1)=0$

$\therefore \cos \theta \ne \frac{-5}{2}$ since range of $[-1,1]$ and $\cos \theta =\frac{1}{2}$

$\therefore \theta =\frac{\pi }{6},2\pi -\frac{\pi }{6}$

$\therefore \theta =\frac{\pi }{6},\frac{12\pi -\pi }{6}$

$\therefore \theta =\frac{\pi }{6},\frac{11\pi }{6}$