## Algebra and Trigonometry 10th Edition

$x=\frac{5\pi}{3}+2n\pi$ $x=\pi+2n\pi$ where $n$ is an integer.
$sin\frac{x}{2}+cos~x=0$ $sin\frac{x}{2}=-cos~x$ $\pm\sqrt {\frac{1-cos~x}{2}}=-cos~x~~$ (square both sides) $\frac{1-cos~x}{2}=cos^2x$ $2~cos^2x+cos~x-1=0$ $2~cos^2x+2~cos~x-cos~x-1=0$ $2~cos~x(cos~x+1)-(cos~x+1)=0$ $(2~cos~x-1)(cos~x+1)=0$ $cos~x=\frac{1}{2}$ or $cos~x=-1$ $cos~x=\frac{1}{2}$: $x=\frac{\pi}{3}$ or $x=\frac{5\pi}{3}$ Testing the solutions: $sin\frac{\frac{\pi}{3}}{2}+cos~\frac{\pi}{3}=\frac{1}{2}+\frac{1}{2}\ne0$ $sin\frac{\frac{5\pi}{3}}{2}+cos~\frac{5\pi}{3}=-\frac{1}{2}+\frac{1}{2}=0$ $cos~x=-1$: $x=\pi$ Testing the solution: $sin\frac{\pi}{2}+cos~\pi=1-1=0$ The period of $cos~x$ is $2\pi$. The general solutions are: $x=\frac{5\pi}{3}+2n\pi$ $x=\pi+2n\pi$ where $b$ is an integer.