Solve the following trigonometric equation #6sin^2x-3sin^2 2x +cos^2x=0# ?

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Answer 1 · 2018-07-11T19:14:19

-5/6pi,5/6pi,-pi/6,pi/6 (+2kpi)# or #-arccos(-sqrt(2/3)),arccos(-sqrt(2/3)),-arccos(sqrt(2/3)),arccos(sqrt(2/3)) (+2kpi)#

We have to solve

#6sin^2(x)-3sin^2(2x)+cos^2(x)=0#
note that #sin(2x)=2sin(x)cos(x)#
and #sin^2(x)+cos^2(x)=1#

so we get

#6(1-cos^2(x))-12(cos^2(x)-cos^4(x))+cos^2(x)=0#
simplifying and cobinging like Terms:

#12cos^4(x)-17cos^2(x)+6=0#

substituting

#t=cos^2(x)#

we get

#12t^2-17t+6=0

#t^2-17/12t+1/2=0#

solving this with the quadratic Formula we get

#t_(1,2)=17/24pmsqrt((17/24)-1/2)#
note that #(17/24)^2-1/2=1/576=1/24^2#

#t_(1,2)=17/24pm1/24#

so we get

#cos^2(x)=3/4#

or

#cos^2(x)=2/3#