Trigonometric Equations

Solve: $\sin^2 \theta - 2 \cos \theta + \dfrac{1}{4} = 0$

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

i.e. $\;$ $1 - \cos ^2 \theta - 2 \cos \theta + \dfrac{1}{4} = 0$

i.e. $\;$ $\cos ^2 \theta + 2 \cos \theta - \dfrac{5}{4} = 0$

i.e. $\;$ $4 \cos^2 \theta + 8 \cos \theta - 5 = 0$

i.e. $\;$ $4 \cos^2 \theta + 10 \cos \theta - 2 \cos \theta - 5 = 0$

i.e. $\;$ $2 \cos \theta \left(2 \cos \theta - 1\right) + 5 \left(2 \cos \theta - 1\right) = 0$

i.e. $\;$ $\left(2 \cos \theta + 5\right) \left(2 \cos \theta - 1\right) = 0$

i.e. $\;$ $\cos \theta = - \dfrac{5}{2}$ $\;$ or $\;$ $\cos \theta = \dfrac{1}{2}$

$\because \;$ $-1 \leq \cos \theta \leq 1$ $\;$ $\therefore \;$ $\cos \theta = - \dfrac{5}{2}$ is not a valid solution.

Principal value of $\cos \theta$ is in $\left[0, \pi\right]$ $\;$ i.e. $\;$ in first or second quadrants.

$\therefore \;$ $\cos \theta = \dfrac{1}{2} = \cos \left(\dfrac{\pi}{3}\right)$ $\;\;\;$ $\left(\because \;\cos \theta = +ve\right)$

$\implies$ $\theta = 2 n \pi \pm \dfrac{\pi}{3}, \;\;\; n \in Z$