How do you solve the following equation cos2x+cosx=0 in the interval [0, 2pi]?

1 Answer
Gió
Jul 4, 2016

I found:
x=pi
x=pi/3 and (5pi)/3

Explanation:

We can use trigonometric identities to change all into cos as:
cos^2(x)-sin^2(x)+cos(x)=0
and:
cos^2(x)-1+cos^2(x)+cos(x)=0
2cos^2(x)+cos(x)-1=0
We can solve this using the Quadratic Formula as in a second degree equation in cos(x);
we can write cos(x)=t
and we get:
2t^2+t-1=0
t_(1,2)=(-1+-sqrt(1+8))/4=(-1+-3)/4
t_1=-1
t_2=1/2
but: t=cos(x) so we have:
cos(x)=-1 when x=pi
cos(x)=1/2 when x=pi/3 and (5pi)/3

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