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
\[\begin{align}
  & \left( \dfrac{3\sin x-\sin 3x}{4} \right)\cos 3x+\left( \dfrac{3\cos x+\cos 3x}{4} \right)\sin 3x+\dfrac{3}{8}=0 \\
 & \dfrac{1}{4}\left[ 3\sin x\cos 3x-\sin 3x\cos 3x+3\cos x\sin 3x+\cos 3x\sin 3x \right]+\dfrac{3}{8}=0 \\
\end{align}\]
\[\begin{align}
  & \dfrac{3}{4}\left[ \sin x\cos 3x+\cos x\sin 3x \right]=-\dfrac{3}{8} \\
 & \Rightarrow \left[ \sin x\cos 3x+\cos x\sin 3x \right]=-\dfrac{1}{2}\to (4) \\
\end{align}\]
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
\[\begin{align}
  & \sin \left( x+3x \right)=-\dfrac{1}{2} \\
 & \sin 4x=-\dfrac{1}{2}\to (6) \\
\end{align}\]
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
\[\begin{align}
  & 4x=n\pi +{{\left( -1 \right)}^{n}}\left( -\dfrac{\pi }{6} \right) \\
 & 4x=n\pi -{{\left( -1 \right)}^{n}}\dfrac{\pi }{6}\text{;}n\in z. \\
\end{align}\]
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.