Been having problem with a few questions from my review for solving trig equations

$\displaystyle

Cos2X(Sin2x-1)=0

$

and

$\displaystyle

2Cos^2x-cosx-1=0

$

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- Feb 14th 2010, 09:29 AMnightrider456Trigonometric equations
Been having problem with a few questions from my review for solving trig equations

$\displaystyle

Cos2X(Sin2x-1)=0

$

and

$\displaystyle

2Cos^2x-cosx-1=0

$ - Feb 14th 2010, 09:37 AMskeeter
$\displaystyle \cos(2x)[\sin(2x)-1]=0$

set each factor equal to 0 and solve ...

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

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

$\displaystyle \sin(2x) = 1$

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

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

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

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