## Precalculus (6th Edition) Blitzer

The solution is $\underline{x=0,\frac{2\pi }{3},\pi ,\frac{4\pi }{3}}.$
We have to solve $\sin 2x+\sin x=0$: \begin{align} & \sin 2x+\sin x=0 \\ & 2\sin x\cos x+\sin x=0 \end{align} Factor: $\sin x\left( 2\cos x+1 \right)=0$ So, from the above result, it can be concluded that, $\sin x=0$ And the other value will be: $2\cos x+1=0$ With the general formula: \begin{align} & \sin x=0 \\ & x=n\pi \end{align} Compute the value of $\cos x$. \begin{align} & 2\cos x+1=0 \\ & 2\cos x=-1 \\ & \cos x=\frac{-1}{2} \end{align} So, in the interval $\left[ 0,2\pi \right),$ the cosine function is $\frac{-1}{2}$ at $\frac{2\pi }{3}\text{ and }\frac{4\pi }{3}$ according to the trigonometric table. \begin{align} & \cos x=\frac{-1}{2} \\ & x=\frac{2\pi }{3} \end{align} The other value is: \begin{align} & \cos x=\frac{-1}{2} \\ & x=\frac{4\pi }{3} \end{align} Since the period of the cosine function is $2\pi$, the general solution of the equation is: $x=\frac{2\pi }{3}+2n\pi \,\text{ and }\frac{4\pi }{3}+2n\pi$ To get different solutions, $\text{put }n=0,1,2,3\ldots$ \begin{align} & \text{For }n=0, \\ & x=n\pi \\ & =0 \end{align} So, another value is: \begin{align} & x=\frac{2\pi }{3}+2n\pi \\ & =\frac{2\pi }{3} \end{align} Other value of x is: \begin{align} & x=\frac{4\pi }{3}+2n\pi \\ & =\frac{4\pi }{3} \end{align} Substitute the value of n: \begin{align} & \text{For }n=1, \\ & x=n\pi \\ & =\pi \end{align} When putting the other values of $n$, the solution becomes outside the interval $\left[ 0,2\pi \right).$ Thus, the exact solution of the equation $\sin 2x+\sin x=0$ on the interval $\left[ 0,2\pi \right)$ is $\underline{x=0,\frac{2\pi }{3},\pi ,\frac{4\pi }{3}}.$