SSC CHSL Model Paper 13

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Question : 61 of 100

Marks: +1, -0

When

2\sin^2θ = 3 \cosθ

, and

0 ≤ θ ≤ 90°

, then θ = ?

45°
30°
90°
60°

Solution:

2\sin^2θ=3\cosθ

2(1-\cos^2θ)=3\cosθ

⇒2-2\cos^2θ=3\cosθ

⇒2\cos^2θ+3\cosθ-2=0

⇒2\cos^2θ+4\cosθ-\cosθ-2=0

⇒2\cosθ(\cosθ+2)

-1(\cosθ+2)=0

⇒(\cosθ+2)(2\cosθ-1)=0

⇒2\cosθ=1

⇒\cosθ=1/2=\cos60°

⇒θ=60°

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