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#.

How do you solve cos^3x+cos^2x-cosx=1cos^3x+cos^2x-cosx=1 for 0<=x<=2pi0<=x<=2pi?