Trigonometric equation

**tradar** · December 15th 2008, 11:15 AM

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

2cos(squared) xtanx-tanx=0

**nzmathman** · December 15th 2008, 11:22 AM

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

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

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

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

To solve the equation, find the solutions for $\cos 2x = 0 \,\,\,\mbox{and} \,\,\,\tan x = 0 \,\,\,\,\mbox{in the interval} \,\,\,\,0 \leq x \leq 2\pi$

**masters** · December 15th 2008, 12:06 PM

Originally Posted by tradar

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

2cos(squared) xtanx-tanx=0

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

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

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

$2\cos^2 x=1$

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

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

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

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

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

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