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Q.
Solutions of the equation $ \sin^2 \theta -\cos \,\theta = \frac{1}{4}, 0 \leq\theta\leq2\pi $
Trigonometric Functions
Solution:
$\sin^2\,\theta- \cos\,\theta=\frac{1}{4}$
$\Rightarrow \,4(1-\cos^2\,\theta)-4\,\cos\,\theta=1$
$\Rightarrow \:\: \,4\,\cos^2\theta+4\,\cos\,\theta-3=0$
$\Rightarrow \,(2\,cos\,\theta+3)(2\,\cos\,\theta-1)=0$
$\Rightarrow \,\cos\,\theta=-\frac{3}{2}$ or $\cos\,\theta=\frac{1}{2}$
But $\,\cos\,\theta=-\frac{3}{2}$ is not possible
$\therefore \,\cos\,\theta=\frac{1}{2}=\cos\,\frac{\pi}{3}$ or $\cos\,\left(2\pi-\frac{\pi}{3}\right)$
$=\cos\,\frac{\pi}{3}$ or $\cos\,\frac{5\,\pi}{3}(0\,\leq\,\theta\,\leq\,2\pi)$
$\therefore \,\theta=\frac{\pi}{3},\frac{5\pi}{3}$