If 0 < θ ≤ 90°, solve for 'θ' where

Correct option is C

Given:

​$$\cos^2\theta - 3 \cos\theta + 2 = 2 \sin^2\theta$$​​

Formula Used:

​$$\sin^2\theta = 1 - \cos^2\theta$$​​
Solution:

​​$$\cos^2\theta - 3 \cos\theta + 2 = 2 \sin^2\theta \\\cos^2\theta - 2 \sin^2\theta - 3 \cos\theta + 2 = 0 \\\cos^2\theta - 2(1 - \cos^2\theta) - 3 \cos\theta + 2 = 0 \\\cos^2\theta - 2 + 2\cos^2\theta - 3 \cos\theta + 2 = 0 \\3\cos^2\theta - 3 \cos\theta = 0 \\3\cos\theta(\cos\theta - 1) = 0 \\3\cos\theta = 0 \text{ and } \cos\theta - 1 = 0 \\\cos\theta = 0 \text{ and } \cos\theta = 1 \\\cos\theta = \cos 90^\circ \text{ and } \cos \theta = cos 0\degree \\\theta = 90^\circ and \theta =0\degree \\\text{according to given option } \theta = 90^\circ$$​​