# How do you solve cosxtanx=1/2?

Dec 9, 2016

$\frac{\pi}{6} , \frac{5 \pi}{6}$

#### Explanation:

$\cos x . \tan x = \frac{1}{2}$
$\cos x \frac{\sin x}{\cos x} = \frac{1}{2}$
Divide by cos x, under condition $\implies$ cos x diff. to zero,
or x diff. to $\frac{\pi}{2} , \frac{3 \pi}{2}$
$\sin x = \frac{1}{2}$
Use trig table of special arcs and unit circle $\implies$
$\sin x = \frac{1}{2}$ $\implies$ arc $x = \frac{\pi}{6}$ , and arc $x = \frac{5 \pi}{6}$
$x = \frac{\pi}{6} + 2 k \pi$
$x = \frac{5 \pi}{6} + 2 k \pi$

Dec 9, 2016

$x = {30}^{\circ}$ or $\frac{\pi}{6}$ and ${150}^{\circ}$ or $\frac{5 \pi}{6}$

#### Explanation:

Using the trigonometric identity, $\textcolor{red}{\tan x} = \frac{\textcolor{g r e e n}{\sin x}}{\textcolor{b l u e}{\cos x}}$,

this question can be written as $\textcolor{b l u e}{\cos x} \cdot \frac{\textcolor{g r e e n}{\sin x}}{\textcolor{b l u e}{\cos x}} = \frac{1}{2}$

$\textcolor{w h i t e}{\text{XXXXXXXXXXXXXXXX}} \textcolor{b l u e}{\cancel{\cos x}} \cdot \frac{\textcolor{g r e e n}{\sin x}}{\textcolor{b l u e}{\cancel{\cos x}}} = \frac{1}{2}$

$\textcolor{w h i t e}{\text{XXXXXXXXXXXXXXXX}} \textcolor{g r e e n}{\sin} x = \frac{1}{2}$

Note: 'N.D.' means Not Defined. For example, $\frac{a}{0}$ is not defined, where $a$ is any non-zero number like 1, 4, 647 etc. But, $\frac{0}{a} = 0$.

As you can see in this table, we get the value of $x$ as ${30}^{\circ}$ or $\frac{\pi}{6}$.

This is called a 'Unit Circle'.

The values in brackets are ($\cos$,$\sin$).
$\sin$ is positive in the first and second quadrants. Since we need $+ \frac{1}{2}$, we consider these quadrants only.

${30}^{\circ}$ is the value of $x$ in the first quadrant.
To get the second value of $x$(in second quadrant), we subtract $\frac{\pi}{6}$ from $\pi$.
$\pi - \frac{\pi}{6} = \frac{6 \pi}{6} - \frac{\pi}{6} = \frac{5 \pi}{6} \mathmr{and} {150}^{\circ}$.

Also note, here $\pi$ is $\pi$ radians, which is equal to ${180}^{\circ}$.
Check out this video for more understanding:
Intro to Arcsin