SSC CHSL 20 Dec 2015 Shift 2 Paper

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Question : 145 of 200

Marks: +1, -0

If

2cos θ − sin θ =1/√2,(0°≤θ≤90°)

the value of 2 sin θ +cosθ is

√2/3
$√2$
$3/√2$
$1/√2$

Solution:

Expression :

2cos θ − sin θ =1/√ 2

Squaring both sides, we get :
⇒

(2cosθ − sin θ )^2 =(1/√ 2)^2

⇒

4cos ^2 θ + sin ^2 θ

− 4sin θ.cos θ = 1/2

⇒

4cos^ 2 θ + (1-cos^ 2 θ)

− 4sin θ.cos θ = 1/2

⇒

3cos^2θ-4sinθ.cosθ=-1/2

------(i)
To find :

2sin θ + cos θ = x

⇒

x^ 2 = 4sin^ 2 θ + cos^ 2 θ

+ 4sinθ.cos θ

⇒

x ^2 = 4(1 − cos^ 2 θ) + cos^ 2 θ

+ 4sin θ.cos θ

⇒

x^ 2 = 4 − 3cos^ 2 θ + 4sin θ.cos θ

⇒

x^ 2 = 4 − (3cos^ 2 θ − 4sin θ.cos θ)

⇒

x^ 2 = 4 -({-1}/2)

[Using equation (i)]
⇒

x^ 2 = {8+1}/2=9/2

⇒

x=√{9/2}=3/√2

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