Answer
The values of x are $\frac{\pi }{6},\frac{5\pi }{6},\,\text{ and }\,\frac{3\pi }{2}$.
Work Step by Step
$\cos 2x-\sin x=0,\,0\le x\,<2\pi $ …… (1)
Substituting the value of $\cos 2x$ as $\cos 2x=1-2{{\sin }^{2}}x$ in equation (1) we get,
$\begin{align}
& 1-2{{\sin }^{2}}x-\sin x=0 \\
& -2{{\sin }^{2}}x-\sin x+1=0
\end{align}$
Multiplying the above equation by $-1$ we get
$\begin{align}
& -2{{\sin }^{2}}x-\sin x+1=0 \\
& 2{{\sin }^{2}}x+\sin x-1=0
\end{align}$
We can also express the equation as below:
$\begin{align}
& 2{{\sin }^{2}}x+\sin x-1=0 \\
& 2{{\sin }^{2}}x+2\sin x-\sin x-1=0
\end{align}$
Now, factorize the equation.
$\begin{align}
& 2\sin x\left( \sin x+1 \right)-1\left( \sin x+1 \right)=0 \\
& \left( 2\sin x-1 \right)\left( \sin x+1 \right)=0
\end{align}$
Solve for x.
$2\sin x-1=0$ or $\sin x+1=0$
$2\sin x=1$ $\sin x=-1$
$\sin x=\frac{1}{2}$ $x=\frac{3\pi }{2}$
$x=\frac{\pi }{6}$
In the range $0\le x\,<2\pi $, the values of x for $2\sin x-1=0$ are $x=\pi -\frac{\pi }{6}=\frac{5\pi }{6}$.
