# Question #00b38

Apr 13, 2016

$\cos 3 x = \frac{1}{2}$
$3 x = {\cos}^{-} 1 \left(\frac{1}{2}\right)$
$3 x = \pm \frac{\pi}{3} + 2 \pi n$
$x = \pm \frac{\pi}{9} + \frac{2}{3} \pi n$

Apr 14, 2016

$x = {20}^{\circ} , {100}^{\circ} , {140}^{\circ} , {220}^{\circ} , {260}^{\circ} , {340}^{\circ}$

#### Explanation:

We can first divide both side of equation by $2$ to eliminate the value of $2$ on $2 \cos 3 x$;

$\frac{2 \cos 3 x}{2} = \frac{1}{2}$

$\cos 3 x = \frac{1}{2}$

The fact that the trigonometry equation of $\cos 3 x$ has positive value means that it is located in Quadrant 1 and Quadrant 4

$x$ initially is in the range from $0$ until $2 \pi$ or $360$ and expressed as;

$x \in \left[{0}^{\circ} , {360}^{\circ}\right]$

But since $x$ now is in the form of $3 x$ we must multiply all the equation in $x \in \left[0 , {360}^{\circ}\right]$ with $3$ and we get;

$3 x \in \left[{0}^{\circ} , {1080}^{\circ}\right]$

The trigonometry equation now gives us a range of
Quadrant 1, 4, 5, 8, 9, and 12 .

Multiply both side in $\cos 3 x = \frac{1}{2}$ with ${\cos}^{-} 1$ to cancel out $\cos$ in $\cos 3 x$;

$\cancel{{\cos}^{-} 1} \cancel{\cos} 3 x = {\cos}^{-} 1 \left(\frac{1}{2}\right)$

$3 x = {60}^{\circ}$

And since $3 x$ is present in Quadrant 1, 4, 5, 8, 9, and 12 ;

$3 x = {60}^{\circ} , \left({360}^{\circ} - {60}^{\circ}\right) , \left({360}^{\circ} + {60}^{\circ}\right) , \left({720}^{\circ} - {60}^{\circ}\right) , \left({720}^{\circ} + {60}^{\circ}\right) , \left({1080}^{\circ} - {60}^{\circ}\right)$

$3 x = {60}^{\circ} , {300}^{\circ} , {420}^{\circ} , {660}^{\circ} , {780}^{\circ} , {1020}^{\circ}$

Divide both side by $3$ and we get the very final answer of all values of $x$;

$\frac{3 x}{3} = {60}^{\circ} / 3 , {300}^{\circ} / 3 , {420}^{\circ} / 3 , {660}^{\circ} / 3 , {780}^{\circ} / 3 , {1020}^{\circ} / 3$

$x = {20}^{\circ} , {100}^{\circ} , {140}^{\circ} , {220}^{\circ} , {260}^{\circ} , {340}^{\circ}$