A) zero
B) 1
C) 2
D) 3
Correct Answer: A
Solution :
\[1+\sin x\cdot {{\sin }^{2}}\frac{x}{2}=0\] \[\Rightarrow \] \[2+2\sin x\cdot {{\sin }^{2}}\frac{x}{2}=0\] \[\Rightarrow \] \[2+\sin x(1-\cos x)=0\] \[\Rightarrow \] \[4+2\sin x(1-\cos x)=0\] \[\Rightarrow \] \[4+2\sin x-\sin 2x=0\] \[\Rightarrow \] \[\sin 2x=2\sin x+4\] Above is not possible for any value of\[x\]as LHS has maximum value 1 and RHS has value 2. Hence, there is no solution.
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