# Trigonometric equation

• Dec 15th 2008, 11:15 AM
Trigonometric equation
Solve the trigonometric equation for the exact values of x, 0<x<2Pi

2cos(squared) xtanx-tanx=0
• Dec 15th 2008, 11:22 AM
nzmathman
$\displaystyle 2\cos^2 x \tan x - \tan x = 0$

$\displaystyle \Rightarrow (\tan x)(2\cos^2 x - 1) = 0$

$\displaystyle \Rightarrow (\tan x)(\cos 2x) = 0$

(identity $\displaystyle 2\cos^2 x - 1 = \cos 2x$)

To solve the equation, find the solutions for $\displaystyle \cos 2x = 0 \,\,\,\mbox{and} \,\,\,\tan x = 0 \,\,\,\,\mbox{in the interval} \,\,\,\,0 \leq x \leq 2\pi$
• Dec 15th 2008, 12:06 PM
masters
Quote:

Solve the trigonometric equation for the exact values of x, 0<x<2Pi

2cos(squared) xtanx-tanx=0

$\displaystyle 2\cos^2x \tan x-\tan x=0$

$\displaystyle \tan x(2\cos^2x-1)=0$

$\displaystyle \tan x=0 \ \ or \ \ 2\cos^x-1=0$

$\displaystyle 2\cos^2 x=1$

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

$\displaystyle \cos x=\pm \sqrt{\frac{1}{2}}=\pm \frac{\sqrt{2}}{2}$

$\displaystyle \tan x=0 \ \ at \ \ \{0, \pi\}$

$\displaystyle \cos x = \pm \frac{\sqrt{2}}{2} \ \ at \ \ \left\{\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}\right\}$

Solution set=$\displaystyle \left\{0, \pi ,\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}\right\}$