SSC CHSL 19 Apr 2021 Shift 3 Paper

Section: Quantitative Aptitude

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Question : 53 of 100

Marks: +1, -0

Solve the following equation and find

θ

.

2 √{3} \sin ^{2} θ+\cos θ-√{3}=0

, where

θ

is an acute angle.

$30^{∘}$
$45^{∘}$
$60^{∘}$
$15^{∘}$

Solution:

Given,

2 √{3} \sin ^{2} θ+\cos θ-√{3}=0

⇒ 2 √{3}(1-\cos ^{2} θ)+\cos θ-√{3}=0

{[}\;{[ ∵ \sin ^{2} θ+\cos ^{2} θ=1] }

⇒ 2 √{3}-2 √{3} \cos ^{2} θ+\cos θ-√{3}=0

⇒ 2 √{3} \cos ^{2} θ-\cos θ-√{3}=0

Let

\cos θ=x

Now,

2 √{3} x^{2}-x-√{3}=0

⇒ 2 √{3} x^{2}-3 x+2 x-√{3}=0

⇒ √{3} x(2 x-√{3})+1(2 x-√{3})=0

∴ x=\;{√{3}}/{2}

and

\;{-1}/{√{3}}

So,

\cos θ=\;{√{3}}/{2}=\cos 30^{∘}

\;\; [

since,

θ

is an acute angle.]

∴ θ=30^{∘}

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