MHB Solve cos 6x=(1/2) for principal values in degree

[blake1]

cos 6x=(1/2)

 

[skeeter]

cos 6x=(1/2)

note $\cos{\theta} = \dfrac{1}{2}$ at $\theta = \dfrac{\pi}{3} \text{ and } \dfrac{5\pi}{3}$

$0 \le x < 2\pi \implies 0 \le 6x < 12\pi$

$\cos(6x) = \dfrac{1}{2} \implies 6x = \dfrac{\pi}{3} \, , \, \dfrac{5\pi}{3} \, , \, \dfrac{7\pi}{3} \, , \, \dfrac{11\pi}{3} \, , \, \dfrac{13\pi}{3} \, , \, \dfrac{17\pi}{3}\, , \, \dfrac{19\pi}{3} \, , \, \dfrac{23\pi}{3} \, , \, \dfrac{25\pi}{3} \, , \, \dfrac{29\pi}{3} \, , \, \dfrac{31\pi}{3} \, , \, \dfrac{35\pi}{3}$

$x = \dfrac{\pi}{18} \, , \, \dfrac{5\pi}{18} \, , \, \dfrac{7\pi}{18} \, , \, \dfrac{11\pi}{18} \, , \, \dfrac{13\pi}{18} \, , \, \dfrac{17\pi}{18}\, , \, \dfrac{19\pi}{18} \, , \, \dfrac{23\pi}{18} \, , \, \dfrac{25\pi}{18} \, , \, \dfrac{29\pi}{18} \, , \, \dfrac{31\pi}{18} \, , \, \dfrac{35\pi}{18}$

 

[HOI]

Skeeter's answer is, of course, in radians. To get the answer in degrees remember that \pi radians is 180 degrees. That is, \frac{180}{\pi}= 1 so \frac{\pi}{18} radians is the same as \frac{\pi}{18}\frac{180}{\pi}= 10 degrees.