# Trigonometric equations

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• February 14th 2010, 09:29 AM
nightrider456
Trigonometric equations
Been having problem with a few questions from my review for solving trig equations

$
Cos2X(Sin2x-1)=0
$

and

$
2Cos^2x-cosx-1=0
$
• February 14th 2010, 09:37 AM
skeeter
Quote:

Originally Posted by nightrider456
Been having problem with a few questions from my review for solving trig equations

$
Cos2X(Sin2x-1)=0
$

and

$
2Cos^2x-cosx-1=0
$

$\cos(2x)[\sin(2x)-1]=0$

set each factor equal to 0 and solve ...

$\cos(2x) = 0$

$2x = \frac{\pi}{2}$ , $2x = \frac{3\pi}{2}$ , ...

$\sin(2x) = 1$

$2x = \frac{\pi}{2}$ , $2x = \frac{5\pi}{2}$ , ...

$2\cos^2{x}-\cos{x}-1=0$

$(2\cos{x} + 1)(\cos{x} - 1) = 0$

same drill ... set each factor equal to 0 and solve.