How do you find all solutions of the equation in the interval $[0,2pi)$ given $2cos^2x-cosx=1$ ?

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Answer 1 · 2017-02-23T09:54:24

The solutions are $S={0,2/3pi,4/3pi}$

We compare this equation to the quadratic equation

$ax^2+bx+c=0$

$2cos^2x-cosx-1=0$

We calculate the discriminant

$Delta=b^2-4ac$

$=(-1)^2-4*(2)*(-1)$

$=1+8=9$

As $Delta>0$ , there are 2 real roots

$x=(-b+-sqrtDelta)/(2a)$

$cosx=(1+-sqrt9)/(4)$

$cosx=(1+3)/4=1$ and $cosx=(1-3)/4=-1/2$

$x=0$ and $x=2/3pi$ and $x=4/3pi$

How do you find all solutions of the equation in the interval [0,2pi)[0,2pi) given 2cos^2x-cosx=12cos^2x-cosx=1?