# Question #1560d

Dec 31, 2016

$\cos 2 \theta - \sin \theta = 0$

$\implies \cos 2 \theta = \sin \theta$

$\implies \cos 2 \theta = \cos \left(\frac{\pi}{2} - \theta\right)$

$\implies 2 \theta = 2 n \pi \pm \left(\frac{\pi}{2} - \theta\right)$

when

$2 \theta = 2 n \pi + \left(\frac{\pi}{2} - \theta\right) \text{ where } n \in \mathbb{Z}$

$\implies 3 \theta = 2 n \pi + \frac{\pi}{2}$

$\implies \theta = \frac{2 n \pi}{3} + \frac{\pi}{6} = \left(4 n + 1\right) \frac{\pi}{6}$

when

$2 \theta = 2 n \pi - \left(\frac{\pi}{2} - \theta\right) \text{ where } n \in \mathbb{Z}$

$\implies \theta = 2 n \pi - \frac{\pi}{2} = \left(4 n - 1\right) \frac{\pi}{2}$

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Alternative

$\cos 2 \theta - \sin \theta = 0$

$\implies 1 - 2 {\sin}^{2} \theta - \sin \theta = 0$

$\implies 2 {\sin}^{2} \theta + \sin \theta - 1 = 0$

$\implies 2 {\sin}^{2} \theta + 2 \sin \theta - \sin \theta - 1 = 0$

$\implies 2 \sin \theta \left(\sin \theta + 1\right) - \left(\sin \theta + 1\right) = 0$

$\implies \left(\sin \theta + 1\right) \left(2 \sin \theta - 1\right) = 0$

So

$\sin \theta + 1 = 0$

$\implies \sin \theta = - 1 = \sin \left(- \frac{\pi}{2}\right)$

$\theta = n \pi - {\left(- 1\right)}^{n} \frac{\pi}{2} \text{ where } n \in \mathbb{Z}$

Again

$2 \sin \theta - 1 = 0$

$\implies \sin \theta = \frac{1}{2} = \sin \left(\frac{\pi}{6}\right)$

$\theta = n \pi + {\left(- 1\right)}^{n} \frac{\pi}{6} \text{ where } n \in \mathbb{Z}$