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#.