Answer
The solutions of the equation are $\frac{\pi }{2}$ , $\frac{3\pi }{2}$ , $\frac{7\pi }{12}$ , $\frac{11\pi }{12}$ , $\frac{19\pi }{12}$ , $\frac{23\pi }{12}$.
Work Step by Step
Solve the provided the equation, by identity $\sin \left( \alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta $.
$\begin{align}
& 0=\sin 3x+\sin x+\cos x \\
& 0=\sin \left( 2x+x \right)+\sin x+\cos x \\
& 0=\sin \left( 2x \right)\cdot \cos x+\cos \left( 2x \right)\sin x+\sin x+\cos x \\
& 0=\left[ \sin \left( 2x \right)\cdot \cos x+\cos x \right]+\left[ \cos \left( 2x \right)\sin x+\sin x \right]
\end{align}$
And take out common terms
$\begin{align}
& 0=\cos x\left( \sin 2x+1 \right)+\sin x\left( \cos 2x+1 \right) \\
& 0=\cos x\left( \sin 2x+1 \right)+\sin x\left( 2{{\cos }^{2}}x-1+1 \right) \\
& 0=\cos x\left( \sin 2x+1 \right)+\sin x\cdot 2{{\cos }^{2}}x \\
& 0=\cos x\sin 2x+\cos x+2\sin x\cos x\cos x
\end{align}$
Simplified,
$\begin{align}
& 0=\sin 2x\cos x+\cos x+\sin 2x\cdot cosx \\
& 0=2\sin 2x\cdot cosx+\cos x \\
& 0=\cos x\left( 2\sin 2x+1 \right)
\end{align}$
Now, $\cos x=0$. It implies,
$x=\frac{\pi }{2}$ , $\frac{3\pi }{2}$
Also,
$\begin{align}
& 2\sin 2x+1=0 \\
& sin2x=-\frac{1}{2} \\
& sin2x=\sin \left( \frac{7\pi }{6} \right)
\end{align}$
Or
$\sin 2x=\sin \left( \frac{11\pi }{6} \right)$
Or
$\sin 2x=\sin \left( \frac{19\pi }{6} \right)$
Or
$\sin 2x=\sin \left( \frac{23\pi }{6} \right)$
Thus, $x=\frac{7\pi }{12}$ , $\frac{11\pi }{12}$ , $\frac{19\pi }{12}$ , $\frac{23\pi }{12}$
Hence, the solutions of the equation are $\frac{\pi }{2}$ , $\frac{3\pi }{2}$ , $\frac{7\pi }{12}$ , $\frac{11\pi }{12}$ , $\frac{19\pi }{12}$ , $\frac{23\pi }{12}$.
