# Question af5d8

May 2, 2017

a. cos 2x = 0
Unit circle gives:
$2 x = \frac{\pi}{2} + 2 k \pi$ and $2 x = \frac{3 \pi}{2} + 2 k \pi$

$2 x = \frac{\pi}{2} + 2 k \pi$ --> $x = \frac{\pi}{4} + k \pi$
$2 x = \frac{3 \pi}{2} + 2 k \pi$ --> $x = \frac{3 \pi}{4} + k \pi$
Answers for $\left(0 , 2 \pi\right)$:
pi/4; (3pi)/4; (5pi)/4; (7pi)/4

b. $\cos x = \cos 2 x = 2 {\cos}^{2} x - 1$
Solve this quadratic equation for cos x:
$2 {\cos}^{2} x - \cos x - 1 = 0$
Since a + b + c = 0, use shortcut.
The 2 real roots are: cos x = 1 and cos x = c/a = - 1/2
Trig table and unit circle give:
1/ cos x = 1 --> x = 0 and x = 2kpi
2/ $\cos x = - \frac{1}{2}$ --> $x = \pm \frac{2 \pi}{3} + 2 k \pi$
Answers for $\left(0 , 2 \pi\right)$:
2pi; +- (2pi)/3
c. sin 3x = 0
Unit circle gives as solutions:
1/ $3 x = 0 + 2 k \pi$ --> $x = \frac{2 k \pi}{3}$
2/ $3 x = \pi + 2 k \pi$--> $x = \frac{\left(2 k + 1\right) \pi}{3}$
3/ 3x = 2pi + 2kpi --> x = (2kpi)/3
Answers for $\left(0 , 2 \pi\right)$:
(2pi)/3; (4pi)/3; 2pi; pi/3; pi#