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
Answer: Option D
: 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
Answer: Option A
: 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
Answer: Option C
: 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
Answer: Option C
: 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
Answer: Option B
: B

The general solution oftanx=tanα,αϵ(π2,π2) is
x=nπ+α,nϵZ

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