The number of solutions of the equation 1+sinxsin2x2=0, in [−π,π]
A
zero
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B
one
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C
two
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D
three
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Solution
The correct option is A zero Given that, 1+sinxsin2x2=0 ∴1+sinx(1−cosx2)=0 ⇒2+sinx–sinxcosx=0 ⇒sin2x–2sinx=4 This is not possible for any x given the fact that range of sine function is [-1,1].