QJEEAdv14Ma165e: JEE Advanced 2014 Mathematics - Class 11 JEEAdv14Ma165e Solution: For the equation has

$\begin{aligned}& \text { Given, } \sin x+2 \sin 2 x-\sin 3 x=3 \quad \text{... (i)}\\& \Rightarrow \quad-[\sin 3 x-\sin x]+2 \sin 2 x=3\end{aligned}$

We know

$\begin{aligned}& \sin \mathrm{C}-\sin \mathrm{D}=2 \cos \left(\frac{\mathrm{C}+\mathrm{D}}{2}\right) \cdot \sin \left(\frac{\mathrm{C}-\mathrm{D}}{2}\right) \\& \text { and } \sin 2 \theta=2 \sin \theta \cdot \cos \theta \\& \therefore-2 \cos \left(\frac{3 x+x}{2}\right) \cdot \sin \left(\frac{3 x-x}{2}\right)+2 \cdot 2 \sin x \cdot \cos x=3 \\& \Rightarrow \quad-2 \cos 2 x \cdot \sin x+4 \sin x \cdot \cos x=3 \\& \Rightarrow \quad-2 \sin x[\cos 2 x-2 \cos x]=3 \\& \Rightarrow \quad-2 \sin x\left[2 \cos ^2 x-1-2 \cos x\right]=3 \\& \Rightarrow \quad-4 \sin x\left[\cos ^2 x-2 \cdot \frac{1}{2} \cos x+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2-\frac{1}{2}\right]=3\end{aligned}$

$\begin{array}{lr}\Rightarrow & -4 \sin x\left[\left(\cos x-\frac{1}{2}\right)^2-\frac{3}{4}\right]=3 \\\Rightarrow & -4 \sin x\left(\cos x-\frac{1}{2}\right)^2+3 \sin x=3 \\\Rightarrow & 3 \sin x-3=4 \sin x\left(\cos x-\frac{1}{2}\right)^2\end{array}$

Draw the graph of $y=3 \sin x-3$ and $y=\left(\cos x-\frac{1}{2}\right)^2$ in the interval $x \in(0, \pi)$

It is clear that the graphs $y=3 \sin x-3$ and $y=\left(\cos x-\frac{1}{2}\right)$ does not intersect in $x \in(0, \pi)$ Hence, the given equation $\sin x+2 \sin 2 x-\sin 3 x=3$ does not have any solution in the interval $x \in(0, \pi)$

Hint:

(i) Recall the formula $\sin 2 x=2 \sin x \cdot \cos x, \cos 2 x=2 \cos ^2 x-1$ and $\sin \mathrm{C}-\sin \mathrm{D}=2 \cos \left(\frac{\mathrm{C}+\mathrm{D}}{2}\right) \cdot \sin \left(\frac{\mathrm{C}-\mathrm{D}}{2}\right)$

(ii) The number of point of intersection of graphs of $y=f(x)$ and $y=g(x)$ is equal to the number of solution of $f(x)=g(x)$ in the given interval.