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B
one solution
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C
two solutions
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D
three solutions
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Solution
The correct option is A no solution sinx+2sin2x−sin3x=3 ⇒sinx+4sinxcosx−3sinx+4sin3x=3⇒sinx(1+4cosx−3+4sin2x)=3⇒sinx(4sin2x+4cosx−2)=3⇒sinx(4cos2x−4cosx−2)=−3⇒sinx[(2cosx−1)2−3]=−3⇒sinx[3−(2cosx−1)2]=3
Maximum value of 3−(2cosx−1)2 is 3 when cosx=12 but at this value, sinx≠1
Hence, sinx[3−(2cosx−1)2]=3 is not possible.