How do you solve #cos^3x+cos^2x-cosx=1# for #0<=x<=2pi#?

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Answer 1 · 2016-09-26T15:57:16

#x=0, pi and 2pi#.

This is a cubic equation in cos x.

Sum of the coefficients is 0. So, cos x = 1 is a solution.

And so, factorizing,

#(cos x - 1)(cos^2+ 2cos x + 1)=0#

The other factor is (cos x + 1)^2.

It follows that the other two solutions are #cos x= -1, -1#.

In brief, #cos x = +-1#

Easily, #x in [0. 2pi] # are 0, pi and 2pi#.