Chapter 5.5, Problem 16E

To determine

To calculate:Thesolution of the equation $\cos 2x+\sin x=0$ in the interval $\left[0,2\pi \right)$ .

Answer to Problem 16E

The solution set of the equation $\cos 2x+\sin x=0$ in the interval $\left[0,2\pi \right)$ is $\left\{\frac{\pi }{2},\frac{7\pi }{6},\frac{11\pi }{6}\right\}$ .

Explanation of Solution

Given information:

The given equation is $\cos 2x+\sin x=0$ .

Formula used:

Use the double angle formula.

  $\cos 2x=1-2{\sin }^{2}x$

Calculation:

Substitute the value of $\cos 2x$ in the given equation.

  $\cos 2x+\sin x=1-2{\sin }^{2}x+\sin x$

Further simplify.

  $\begin{array}{c}1-2{\sin }^{2}x+\sin x=0\\ 2{\sin }^{2}x-\sin x-1=0\\ 2{\sin }^{2}x-2\sin x+\sin x-1=0\\ \left(2\sin x+1\right)\left(\sin x-1\right)=0\end{array}$

The solutions in the interval $\left[0,2\pi \right)$ , substitute the factor $2\sin x+1=0$ .

  $x=\frac{7\pi }{6},\frac{11\pi }{6}$

The solutions in the interval $\left[0,2\pi \right)$ , substitute the factor $\sin x-1=0$ .

  $x=\frac{\pi }{2}$

The solution set is

  $\left\{\frac{\pi }{2}\right\}$ .

Therefore, the solution set is

  $\left\{\frac{7\pi }{6},\frac{11\pi }{6}\right\}{\displaystyle \cup \left\{\frac{\pi }{2}\right\}=\left\{\frac{\pi }{2},\frac{7\pi }{6},\frac{11\pi }{6}\right\}}$

Verify the solutions using graphical utility:

Draw the graph for the equation $\cos 2x+\sin x=0$ as shown in Figure(1).

  

Figure(1)

From the above, it is clear that $\left\{\frac{\pi }{2},\frac{7\pi }{6},\frac{11\pi }{6}\right\}$ are the solutions.

Therefore, the solution of the equation $\cos 2x+\sin x=0$ in the interval $\left[0,2\pi \right)$$\left\{\frac{\pi }{2},\frac{7\pi }{6},\frac{11\pi }{6}\right\}$ .