Question

2cos2theta + √2sintheta =2​

Answer 1

Step-by-step explanation:

We have,

$2 \cos(2 \theta) + \sqrt{2 \sin(\theta) } = 2$

$= > \sqrt{2 \sin(\theta) } = 2(1 - \cos(2\theta) )$

$= > 2 \sin(\theta) = 16 \sin^{4} (\theta)$

$= >8 \sin^{4} (\theta) - \sin(\theta) = 0$

$= > \sin(\theta) (8 \sin^{3} (\theta) - 1) = 0$

$= > \sin(\theta) (2 \sin(\theta) - 1)(4 \sin^{2} (\theta) + 2 \sin(\theta) + 1) = 0$

$= > \sin(\theta) = 0 \: \: or \: \: \sin(\theta) = \frac{1}{2}$

$= > \theta = n\pi \: \: or \: \: \theta = n\pi + ( - 1)^{n} \frac{\pi}{6}$