How do you find all solutions of the equation in the interval $[0,2pi)$ given $2cos^2x+3cosx=0$ ?

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Answer 1 · 2017-02-10T15:55:22

$x={pi/2, (3pi)/2}$

$2cos^2x+3cosx=0$

$cosx(2cosx+3)=0$

$cosx=0 or 2cosx+3=0$

$x=cos^-1 0 or cosx=-3/2$ -->Not possible

$x=cos^-1 0$

$x={pi/2, (3pi)/2}$

How do you find all solutions of the equation in the interval [0,2pi)[0,2pi) given 2cos^2x+3cosx=02cos^2x+3cosx=0?