Question

Find the general solution of the equation: sin² θ - cos θ = $\frac{1}{4}$.

Answer 1

As we know that

$1 - {cos}^{2} \theta = {sin}^{2}\theta$
so

$1 - {cos}^{2} \theta - cos \: \theta = \frac{1}{4} \\ \\ - {cos}^{2} \theta - cos \: \theta + 1 - \frac{1}{4} = 0 \\ \\ - {cos}^{2} \theta - cos \: \theta + \frac{3}{4} = 0 \\ \\ 4{cos}^{2} \theta + 4cos \: \theta - 3 = 0 \\ \\ 4{cos}^{2} \theta + 6cos \: \theta - 2 \: cos \: \theta - 3 = 0 \\ \\ 2 \: cos \: \theta(cos \: \theta + 3) - 1(cos \: \theta + 3) = 0 \\ \\ (2 \: cos \: \theta - 1)(cos \: \theta + 3) = 0$
$2cos \: \theta - 1= 0 \\ \\ cos \: \theta = \frac{1}{2} \\ \\ \theta = {cos}^{ - 1} ( \frac{1}{2} ) \\ \\ \theta = \frac{\pi}{3}$
we had to discard another value of θ.

So general Solution of equation is

$\theta = ±\frac{\pi}{3} + 2 \: \pi \: k$,where k is any Integer.