Solution assignment 03 Trigonometric equations

Solve the following equations:

$\sin(x+\displaystyle\frac{\pi}{3})=\displaystyle\frac{1}{2}\sqrt{2}$

$\cos(-x)=\displaystyle\frac{1}{2}$

Looking at the first equation we use the table:

$x+\displaystyle\frac{\pi}{3}=\displaystyle\frac{\pi}{4}$

$x=-\displaystyle\frac{1}{12}\pi$

but also:

$x+\displaystyle\frac{\pi}{3}=\pi-\displaystyle\frac{\pi}{4}=\displaystyle\frac{3\pi}{4}$

$x=\displaystyle\frac{3\pi}{4}-\displaystyle\frac{\pi}{3}=\displaystyle\frac{5}{12}\pi$

Because the $\sin$ -function has period $2\pi$ , we find the following solution:

$x=-\displaystyle\frac{\pi}{12}+2k\pi, k=0. \pm1, \pm2, ...$

$x=\displaystyle\frac{5\pi}{12}+2k\pi, k=0, \pm1, \pm2, ...$

For the second equation we have:

$\cos(-x)=\cos(x)$

and thus the equation is:

$\cos(x)=\displaystyle\frac{1}{2}$

and from the table we know that:

$x=\displaystyle\frac{\pi}{3}+2k\pi, k=0, \pm1, \pm2, ...$

or:

$x=-\displaystyle\frac{\pi}{3}+2k\pi, k=0, \pm1, \pm2, ...$