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Question

# Which of the following set of values of x satisfies the equation 22sin2x−3sinx+1+22−2sin2x+3sinx=9 (where n∈Z)

A
nπ+π4
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B
nπ+π6
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C
nππ6
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D
2nπ+π2
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Solution

## The correct options are B nπ+π6 C nπ−π6 D 2nπ+π2 22sin2x−3sinx+1+22−2sin2x+3sinx=9 22sin2x−3sinx+1+23−(2sin2x−3sinx+1)=9 Let 22sin2x−3sinx+1=t ∴t+8t=9⇒t2−9t+8=0⇒t=1,8 When t=1 22sin2x−3sinx+1=1 ⇒2sin2x−3sinx+1=0⇒(2sinx−1)(sinx−1)=0⇒sinx=12,1 When t=8 22sin2x−3sinx+1=8⇒2sin2x−3sinx+1=3⇒2sin2x−3sinx−2=0⇒(2sinx+1)(sinx−2)=0⇒sinx=−12 (∵sinx∈[−1,1]) From both, we get sinx=±12,1⇒x=nπ±π6, x=2nπ+π2,n∈Z

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