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Question

Which of the following set values of x satisfies the equation 2(2sin2x3sinx+1)+2(22sin2x+3sinx)=9

A
x=nπ±π6,nϵI
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B
x=nπ±π3,nϵI
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C
x=nπ,nϵI
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D
x=2nπ+π2,nϵI
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Solution

The correct options are
A x=nπ,nϵI
C x=nπ±π3,nϵI

We can write the above expression as,
22sin2x3sinx+1+2322sin2x3sinx+1=9
Let,
22sin2x3sinx+1=k
So,
k+8k=9


k=9±72=8or1
22sin2x3sinx+1=8=23 or 22sin2x3sinx+1=1=20
2sin2x3sinx+1=3 or 2sin2x3sinx+1=0
2sin2x3sinx2=0 or (2sinx1)(sinx1)=0
sinx=12or2 or sinx=12Orsinx=1
So,
sinx=12,12,sinx=1
So, x=nπ or x=nπ±π3


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