How do you solve the equation #sin^2x=3cos^2x#?

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How do you solve the equation #sin^2x=3cos^2x#?

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Answer 1 · 2018-04-05T06:00:02

#x=pi/3,(2pi)/3,(4pi)/3,(5pi)/3#

Explanation:

#(sinx)^2=3(cosx)^2#
#(sinx)^2=3(1-(sinx)^2)#
#(sinx)^2=3-3(sinx)^2#
#4(sinx)^2=3#
#(sinx)^2=3/4#
#sinx=+-(sqrt3/2)#
#x=pi/3,pi-pi/3,pi+pi/3,(2pi)-pi/3#
#x=pi/3,(2pi)/3,(4pi)/3,(5pi)/3#

If this was in the region #0<=x<=2pi#

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How do you solve the equation #sin^2x=3cos^2x#?

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