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Trigonometrical Equations
hard
For $x \in(0, \pi)$, the equation $\sin x+2 \sin 2 x-\sin 3 x=3$ has
A
infinitely many solutions
B
three solutions
C
one solution
D
no solution
(IIT-2014)
Solution
$\sin x+2 \sin 2 x-\sin 3 x=3 $
$\sin x\left(1+2 \cos x-3+4 \sin ^2 x\right)=3 $
$\left(4 \sin ^2 x+2 \cos x-2\right)=\frac{3}{\sin x} $
$2-4 \cos ^2 x+2 \cos x=\frac{3}{\sin x} $
$\frac{9}{4}-\left(2 \cos x-\frac{1}{2}\right)^2=\frac{3}{\sin x} $
$\text { L.H.S. } \leq \frac{9}{4} \quad \quad \text { R.H.S. } \geq 3$
No solution.
Standard 11
Mathematics
