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

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Answer 1 · 2017-01-15T07:23:52

Solution is ${0,(2pi)/3,(4pi)/3}$

$2cos^2x-cosx-1=0$ can be written as

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

or $2cosx(cosx-1)+1(cosx-1)=0$

or $(2cosx+1)(cosx-1)=0$

$:.$ either $2cosx+1=0$ i.e. $cosx=-1/2$ and in the interval $[0,2pi)$ $x=(2pi)/3$ or $(4pi)/3$

or $cosx-1=$ i.e. $cosx=1$ and in given interval $x=0$ .

Hence solution is ${0,(2pi)/3,(4pi)/3}$

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