How do you find all solution of the equation $4 cos^2x-3=0$ ?

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Answer 1 · 2015-02-17T18:20:26

The answer are:

$x=pi/6+2kpiandx=11/6pi+2kpi$ ,

$x=5/6pi+2kpiandx=7/6pi+2kpi$ .

This is because:

$4cos^2x-3=0rArrcos^2x=3/4rArrcosx=+-sqrt3/2$ .

So:

$cosx=sqrt3/2rArrx=pi/6+2kpiandx=11/6pi+2kpi$ ,

$cosx=-sqrt3/2rArrx=5/6pi+2kpiandx=7/6pi+2kpi$ .

How do you find all solution of the equation 4 cos^2x-3=04 cos^2x-3=0?