Recall that #cos2x=cos^2x-sin^2x#, and apply this to the equation:
#cos^2x-sin^2x+cosx=0#
The objective now is to have this equation in terms of one trigonometric function. Since there is only one sine, it will be easiest to get rid of the sine and have only cosines. Recall that
#sin^2x+cos^2x=1#
#sin^2x=1-cos^2x#
Apply this to the above equation:
#cos^2x-(1-cos^2x)+cosx=0#
#cos^2x-1+cos^2x+cosx=0#
#2cos^2x+cosx-1=0#
Now, this resembles a quadratic equation, but instead of having #ax^2+bx+c=0,# we have #acos^2x+bcosx+c=0#. So, this can be factored just as a quadratic equation can. I'll leave the factoring to you -- use any method you like.
Factored form:
#(2cosx-1)(cosx+1)=0#
Now, we must solve
#2cosx-1=0#
#cosx+1=0#
For #2cosx-1=0:#
#2cosx=1#
#cosx=1/2#, holds true for #x=pi/3+2npi, x=(5pi)/3+2npi# as the period of cosine is #2pi,# meaning all values of cosine repeat every #2pi# units.
For #cosx+1=0:#
#cosx=-1#
#x=(3pi)/2+2npi#
