Question

Find the values of x if $4\sin x.\sin 2x.\sin 4x=\sin 3x$.

Answer 1

Hint: Write 2x as $(3x-x)$ and 4x as $(3x+x)$ . Use the formula $\sin 3x=3\sin x-4{{\sin }^{3}}x$ and $\sin (A+B)\sin (A-B)={{\sin }^{2}}A-{{\sin }^{2}}B$ and transform the given equation as $4\sin x.\sin (3x-x).\sin (3x+x)=\sin 3x$ and solve them further. General solution of $\sin x=0$ is $n\pi$ and $\sin x=\sin \left( \pm \dfrac{\pi }{3} \right)$ is $n\pi \pm \dfrac{\pi }{3}$ .

Complete step-by-step solution -
According to the question, we have the equation $4\sin x.\sin 2x.\sin 4x=\sin 3x$ ……………(1)
Here, we have to find the values of x which satisfies the given equation.
First of all, we have to simplify the given equation.
We can write 2x as $(3x-x)$ and 4x as $(3x+x)$ .
Now, replacing 2x by $(3x-x)$ and 4x by $(3x+x)$ in the equation (1), we get
$4\sin x.\sin 2x.\sin 4x=\sin 3x$
$\Rightarrow 4\sin x.\sin (3x-x)\sin (3x+x)=\sin 3x$ ………………………(2)
We know the formula, $\sin (A+B)\sin (A-B)={{\sin }^{2}}A-{{\sin }^{2}}B$ .
Replacing A by 3x and B by x in the equation $\sin (A+B)\sin (A-B)={{\sin }^{2}}A-{{\sin }^{2}}B$ , we get
$\sin (3x-x)\sin (3x+x)={{\sin }^{2}}3x-{{\sin }^{2}}x$ ……………..(3)
Transforming equation (2) using equation (3), we get
$\Rightarrow 4\sin x({{\sin }^{2}}3x-{{\sin }^{2}}x)=\sin 3x$…………..(4)
Now, expanding $\sin 3x$ , we get
$\sin 3x=\sin (2x+x)=3\sin x-4{{\sin }^{3}}x$ ……………………(5)
Now, using equation (5) we can transform equation (4).

& \Rightarrow 4\sin x({{\sin }^{2}}3x-{{\sin }^{2}}x)=3\sin x-4{{\sin }^{3}}x \\\ & \Rightarrow 4\sin x.{{\sin }^{2}}3x-4{{\sin }^{3}}x=3\sin x-4{{\sin }^{3}}x \\\ & \Rightarrow 4\sin x.{{\sin }^{2}}3x-3\sin x=0 \\\ \end{aligned}$$ $$\Rightarrow \sin x(4{{\sin }^{2}}3x-3)=0$$ $$\sin x=0$$ or $$4{{\sin }^{2}}3x-3=0$$ . Now, taking $$\sin x=0$$ and we know that the general solution of $$\sin x=0$$ is $$n\pi $$. $$\begin{aligned} & \sin x=0 \\\ & \Rightarrow x=n\pi \\\ \end{aligned}$$ Now, taking $$4{{\sin }^{2}}3x-3=0$$ $$\begin{aligned} & 4{{\sin }^{2}}3x-3=0 \\\ & \Rightarrow \sin 3x=\pm \dfrac{\sqrt{3}}{2} \\\ \end{aligned}$$ We know that $$\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}$$ and $$\sin \left( -\dfrac{\pi }{3} \right)=-\dfrac{\sqrt{3}}{2}$$ . We have, $$\sin \left( \pm \dfrac{\pi }{3} \right)=\pm \dfrac{\sqrt{3}}{2}$$ . Now, solving $$\sin 3x=\pm \dfrac{\sqrt{3}}{2}$$ $$\Rightarrow \sin 3x=\sin \left( \pm \dfrac{\pi }{3} \right)$$ $$\begin{aligned} & \Rightarrow 3x=n\pi \pm \dfrac{\pi }{3} \\\ & \Rightarrow x=\dfrac{n\pi }{3}\pm \dfrac{\pi }{9} \\\ \end{aligned}$$ So, $$x=n\pi $$ or $$x=\dfrac{n\pi }{3}\pm \dfrac{\pi }{9}$$. Note: In this question, one can try to simplify the equation $$4\sin x.\sin 2x.\sin 4x=\sin 3x$$ using the formula $$2\sin A.\sin B=\cos (A-B)-\cos (A+B)$$ . One can consider x as A and 4x as B and then apply this formula. But when we apply this formula then we will get cosine terms. It means we will have sine terms and cosine terms in the equation due to which our equation needs to be converted into either sine terms or cosine terms. While conversion, we will face complexity. So, don’t approach this question by this method as it leads to complexity.