Question
For $$x \in \left( {0,\pi } \right),$$ the equation $$\sin x + 2\sin 2x - \sin 3x = 3$$ has
A.
infinitely many solutions
B.
three solutions
C.
one solution
D.
no solution
Answer :
no solution
Solution :
$$\eqalign{
& \sin \,x + 2\sin \,2x - \sin \,3x = 3 \cr
& \Rightarrow \,\,\sin \,x + 4\sin \,x\,\cos \,x - 3\sin \,x + 4\,{\sin ^3}\,x = 3 \cr
& \Rightarrow \,\,\sin \,x\,\left( { - 2 + 2\,\cos \,x + 4\,{{\sin }^2}x} \right) = 3 \cr
& \Rightarrow \,\,\sin \,x\,\left( { - 2 + 2\,\cos \,x\, + 4 - 4\,{{\cos }^2}x} \right) = 3 \cr
& \Rightarrow \,\,2 + 2\,\cos \,x - 4\,{\cos ^2}x = \frac{3}{{\sin \,x}} \cr
& \Rightarrow \,\,2 - \,\,\left( {4\,{{\cos }^2}\,x - 2.\,\,2\,\cos \,x.\frac{1}{2} + \frac{1}{4}} \right)\,\, + \,\frac{1}{4} = \frac{3}{{\sin \,x}} \cr
& \Rightarrow \,\,\frac{9}{4} - {\left( {2\,\cos \,x\, - \frac{1}{2}} \right)^2} = \frac{3}{{\sin \,x}} \cr
& {\text{LHS}}\, \leqslant \,\frac{9}{4}\,{\text{and}}\,{\text{RHS}}\, \geqslant 3 \cr
& \therefore \,\,{\text{Equation}}\,{\text{has}}\,{\text{no}}\,{\text{solution}} \cr} $$