Question

Solve the equation${{\sin }^{3}}x\cos 3x+{{\cos }^{3}}x\sin 3x+\dfrac{3}{8}=0$.

Answer 1

Hint: Get the values of${{\sin }^{3}}x$ and ${{\cos }^{3}}x$from the trigonometric identities of $\sin 3x$and$\cos 3x$. Identities are given as $\sin 3x=3\sin x-4{{\sin }^{3}}x,\text{ }\cos 3x=4{{\cos }^{3}}x-3\cos x$
Simplify it further and again use trigonometric identity of $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B.$General solution of equation $\sin x=\sin y,$is given as$x=n\pi \pm \left( -1 \right)$, where$n\in z$.

Complete step-by-step answer:
Given expression in the problem is
${{\sin }^{3}}x\cos 3x+{{\cos }^{3}}x\sin 3x+\dfrac{3}{8}=0...........(1)$
We know the algebraic identities of$\sin 3x$ and$\cos 3x$are given as
$\sin 3x=3\sin x-4{{\sin }^{3}}x$
$\cos 3x=4{{\cos }^{3}}x-3\cos x$
Hence, we can get the value of${{\sin }^{3}}x$ and ${{\cos }^{3}}x$from both the expression as
{{\sin }^{3}}x=\dfrac{3\sin x-\sin 3x}{4}$$$$\to (2)
{{\cos }^{3}}x=\dfrac{3\cos x+\cos 3x}{4}$$$$\to (3)
Now, put the value of${{\sin }^{3}}x$ and ${{\cos }^{3}}x$from the equation (2) and (3) to the equation (1). So, we get

& \left( \dfrac{3\sin x-\sin 3x}{4} \right)\cos 3x+\left( \dfrac{3\cos x+\cos 3x}{4} \right)\sin 3x+\dfrac{3}{8}=0 \\\ & \dfrac{1}{4}\left[ 3\sin x\cos 3x-\sin 3x\cos 3x+3\cos x\sin 3x+\cos 3x\sin 3x \right]+\dfrac{3}{8}=0 \\\ \end{aligned}$$ $$\begin{aligned} & \dfrac{3}{4}\left[ \sin x\cos 3x+\cos x\sin 3x \right]=-\dfrac{3}{8} \\\ & \Rightarrow \left[ \sin x\cos 3x+\cos x\sin 3x \right]=-\dfrac{1}{2}\to (4) \\\ \end{aligned}$$ Now, we know that the trigonometric identity of$$\sin \left( A+B \right)$$is given as $$\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B\to (5)$$ Hence, we can simplify equation (4) further, with the help of equation (5) as $$\begin{aligned} & \sin \left( x+3x \right)=-\dfrac{1}{2} \\\ & \sin 4x=-\dfrac{1}{2}\to (6) \\\ \end{aligned}$$ We know$$\sin \theta $$ gives the value $$\dfrac{1}{2}$$ at $$\dfrac{\pi }{6}$$ and we know the relation $$\sin \left( -\theta \right)=-\sin \theta $$ as well. So, put $$\theta =\dfrac{\pi }{6}$$ to the expression $$\sin \left( -\theta \right)=-\sin \theta $$, we get $$\sin \left( -\dfrac{\pi }{6} \right)=-\sin \left( \dfrac{\pi }{6} \right)=-\dfrac{1}{2}$$ Hence, the principal value of $$4x$$from the equation (6) will be $$-\dfrac{\pi }{6}$$. Now, we can write the equation (6) as $$\sin 4x=\sin \left( -\dfrac{\pi }{6} \right)$$ Now, we know the general solution of $$\sin x=\sin y$$ is given as $$x=n\pi +{{\left( -1 \right)}^{n}}y;n\in z$$. So, we get general solution as $$\begin{aligned} & 4x=n\pi +{{\left( -1 \right)}^{n}}\left( -\dfrac{\pi }{6} \right) \\\ & 4x=n\pi -{{\left( -1 \right)}^{n}}\dfrac{\pi }{6}\text{;}n\in z. \\\ \end{aligned}$$ Now, divide the whole equation by ‘4’, $$x=\dfrac{n\pi }{4}-{{\left( -1 \right)}^{n}}\dfrac{\pi }{24}$$ Note: Another approach for this question would be that we could put values of $$\sin 3x$$ and$$\cos 3x$$ by their identities. Hence, may get a 6-degree equation related to the given equation. Do not confuse with the formula of $$\sin 3x=3\sin x-4{{\sin }^{3}}x$$ $$\cos 3x=4{{\cos }^{3}}x-3\cos x$$ Do not confuse with the other general equation of$$\sin x=\sin y$$. We need to use principle value for the c. One may connect the whole relation to $$\cos \theta $$as well.