# How do you solve cos x + sin x tan x = 2 over the interval 0 to 2pi?

May 9, 2018

$x = \frac{\pi}{3}$
$x = \frac{5 \pi}{3}$

#### Explanation:

$\cos x + \sin x \tan x = 2$

$\textcolor{red}{\tan x = \frac{\sin x}{\cos x}}$

$\cos x + \sin x \left(\sin \frac{x}{\cos} x\right) = 2$

$\cos x + {\sin}^{2} \frac{x}{\cos} x = 2$

${\cos}^{2} \frac{x}{\cos} x + {\sin}^{2} \frac{x}{\cos} x = 2$

$\frac{{\cos}^{2} x + {\sin}^{2} x}{\cos} x = 2$

$\textcolor{red}{{\cos}^{2} x + {\sin}^{2} x = 1}$

$\textcolor{red}{\text{the phythagrean identity}}$

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

multiply both sides by $\cos x$

$1 = 2 \cos x$

divide both sides by $2$

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

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

from the unit circle $\cos \left(\frac{\pi}{3}\right)$ equals $\frac{1}{2}$

so

$x = \frac{\pi}{3}$

and we know that $\cos$ is positive in the first and fourth quadrant so find an angle in the fourth quadrant that $\frac{\pi}{3}$ is the reference angle of it

so

$2 \pi - \frac{\pi}{3} = \frac{5 \pi}{3}$

so

$x = \frac{\pi}{3} , \frac{5 \pi}{3}$

May 9, 2018

$x = \frac{\pi}{3} \mathmr{and} \frac{5 \pi}{3}$

#### Explanation:

The way I'm checking the other answer is writing my own.

$\cos x + \sin x \tan x = 2$

$\cos x + \sin x \left(\sin \frac{x}{\cos} x\right) = 2$

${\cos}^{2} x + {\sin}^{2} x = 2 \cos x$

$1 = 2 \cos x$

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

There's the cliche triangle, you knew it was coming.

In the range,

$x = \frac{\pi}{3} \mathmr{and} \frac{5 \pi}{3}$

Check:

 cos({5pi}/3) + sin ({5pi}/3) tan({5pi}/3) = 1/2 + -\sqrt{3}/2 cdot {-sqrt{3}//2}/{1//2} = 1/2 + 3/2 = 2 quad sqrt