NIMCET 2013 Question Paper with solutions

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Question : 14 of 120

Marks: +1, -0

The solution of sin x + 1 = cos x such that 0 ≤ x ≤ 2π is

0 , π
0 , π/2
π/2 , 3π/2
0 , 3π/2

Solution:

Given,

\sin x+1=\cos x

⇒\cos x-\sin x=1

Dividing by

√2

on both sides, we get

{\cos x}/√2-{\sin x}/√2=1/√2

⇒1/√2\cos x-1/√2\sin x=1/√2

⇒\cos(x+π/4)=\cosπ/4

∴x+π/4=2nπ±π/4

On putting

n=0

x+π/4=0±π/4

⇒x=0

[∵ taking

+π/4

]
Now, putting n= 1,

x+\;\;{π}/{4}=2×1×π±\;\;{π}/{4}⇒x=2π-\;\;{π}/{4}-\;\;{π}/{4}

⇒x={3π}/4

[∵ taking

-π/4

]

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