## Lakshya Education MCQs

Question: If sinx2=sin β

where π2βπ2 then,
Options:
 A. x=nπ+(−1)n β B. x=2nπ+(−1)n β C. x=nπ+(−1)n 2β D. x=2nπ+(−1)n 2β
: D

sinx2=sinβ,π2βπ2x2=nπ+(1)nβx=2nπ+(1)n2β

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## More Questions on This Topic :

Question 1. Find the general solution of sin x+2=sin x
1.    x=nπ+(−1)n+1π4
2.    x=nπ+(−1)nπ4
3.    x=nπ+(−1)nπ3
4.    x=nπ+(−1)n3π4
: A

sinx+2=sinx2sinx=2sinx=12x=nπ+(1)n(π4)x=nπ+(1)n+1π4
Question 2. What is the principal solution of 4 cos2x8 cos x+3=0 ?
1.    x=π6
2.    x=π4
3.    x=π3
4.    x=π2
: C

4cos2x8cosx+3=04cos2x8cosx+4=1(2cosx2)2=12cosx2=±12cosx=3notpossible2cosx=1cosx=12x=π3
Question 3. Find the principal solution of 3cos2θcos 2θ=1
1.    θ=0
2.    θ=−π2
3.    θ=π2
4.    θ=π
: C

3cos2θcos2θ=13cos2θ(2cos2θ1)=1cos2θ=0cosθ=0θ=π2
Question 4. The general solution of tan x=tan α, α ϵ (π2,π2) is
1.    x=nπ+(−1)nα, n ϵ Z
2.    x=nπ+α, n ϵ Z
3.    x=2nπ±α, n ϵ Z
4.    None of these