How do you find the general solutions for $2 {cos}^{2} x - cos x = 0$ $2 cos^2 x - cos x = 0$ ?

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Answer 1 · 2018-04-08T03:58:59

$x = \frac{π}{2} + n π, x = \frac{π}{3} + 2 n π, x = \frac{5 π}{3} + 2 n π$ $x=pi/2+npi, x=pi/3+2npi, x=(5pi)/3+2npi$

All terms share the cosine in common. We can thus factor out a cosine:

$cos x (2 cos x - 1) = 0$ $cosx(2cosx-1)=0$

We may now solve the following two:

a. $cos x = 0$ $cosx=0$

$x = \frac{π}{2} + n π$ $x=pi/2+npi$

As cosine is zero for $pi/2, (3pi)/2, (5pi)/2,...,pi/2+npi$
$2cosx-1=0$

$2cosx=1$

$cosx=1/2$

$x=pi/3+2npi$

$x=(5pi)/3+2npi$

As cosine is equal to (positive) one half in quadrants I, IV.

How do you find the general solutions for 2 cos^2 x - cos x = 02cos2x−cosx=02 cos^2 x - cos x = 0?