### Home > A2C > Chapter 9 > Lesson 9.3.2 > Problem9-155

9-155.

For each equation, find two solutions $0 \lt x \lt 2$, which make the equation true. No calculator necessary.

1. $\operatorname { cos } x = - \frac { 1 } { 2 }$

$\text{cos}\left(\frac{\pi}{6}\right)=\frac{1}{2}$

What angle has the cosine of negative one half? Think of 30°, 60° right triangles.

x = 120° or 240°

$120\left(\frac{\pi}{180}\right)\,\,\,\,\,\,\,240\left(\frac{\pi}{180}\right)$

Now convert the angles into radians.

$\textit{x}=\frac{2\pi}{3},\,\frac{4\pi}{3}$

1. $\operatorname{tan} x =\frac { \sqrt { 3 } } { 3 }$

See part (a).

1. $\operatorname{sin} x = 0$

If sin(x) = 0, y = 0

$x = 0$

1. $\operatorname{cos} x =\frac { \sqrt { 2 } } { 2 }$

Remember the side ratios of $45°$ right triangles.