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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