Question

14 no. anyone solve. I will mark BRAINLIST

Answer 1

Hey There!
Here, we have to find $\sin \theta$

We can solve it easily as follows:
$2 \sin \theta = 2 - \cos \theta \\ \\ \implies \cos \theta = 2 - 2 \sin \theta \\ \\ \cos \theta = 2(1-\sin \theta) \\ \\ Squaring \\ \\ \implies \cos^2\theta = 4(1-\sin\theta)^2 \\ \\ \implies (1-\sin^2\theta) = 4(1-2\sin\theta + \sin^2\theta) \\ \\ \implies 1-\sin^2\theta = 4 -8\sin\theta + 4 \sin^2\theta \\ \\ \implies 5\sin^2\theta -8\sin\theta + 3 = 0 \\ \\ \implies 5\sin^2\theta -5\sin\theta - 3\sin \theta +3 = 0 \\ \\ \implies 5\sin\theta(\sin\theta-1)-3(\sin\theta-1)=0 \\ \\ \implies (\sin\theta-1)(5\sin\theta-3)=0 \\ \\ \implies \sin\theta-1=0 \, \, \, OR \, \, \, 5\sin\theta-3=0 \\ \\ \implies \boxed{\sin\theta=1} \, \, \, OR \, \, \, \boxed{ \sin\theta = \frac{3}{5}}$


Hope it helps
Purva
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