SSC CGL Tier II 12 Apr 2015 Paper

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Question : 39 of 100

Marks: +1, -0

The value of q (0 ≤ θ ≤ 90°) satisfying

2 sin^2θ = 3 cos θ

is

60°
30°
90°
45°

Solution:

(1)

2 \sin ^{2} \θ=3 \cos \θ

⇒ 2 (1-\cos ^{2} \θ)=3 \cos\θ

⇒ 2-2 \cos ^{2} \θ=3 \cos \θ

⇒ 2 \cos ^{2} \θ+3 \cos\θ-2=0

⇒ 2 \cos ^{2}\θ+4 \cos \θ-\cos\θ-2

=0

⇒ 2 \cos\θ(\cos \θ+2)-1(\cos \θ+

2 )=0

⇒(2 \cos\θ-1)(\cos \θ+2)=0

⇒(2 \cos\θ-1)=0

because

\cos \θ+2 ≠ 0

⇒ 2 \cos\θ=1

⇒\cos \θ= {1}/{2}=\cos 60°

⇒\θ=60°

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