## Precalculus (6th Edition) Blitzer

The values of x are $\frac{\pi }{6},\frac{5\pi }{6},\,\text{ and }\,\frac{3\pi }{2}$.
$\cos 2x-\sin x=0,\,0\le x\,<2\pi$ …… (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 $x=\pi -\frac{\pi }{6}=\frac{5\pi }{6}$.