#6(tanx)^2-4(sinx)^2=1#. Solve x?

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Answer 1 · 2017-10-11T15:14:47

#6(tanx)^2-4(sinx)^2=1#

#=>6(tan^2x)xxcos^2x-4sin^2x xxcos^2x=cos^2x#

#=>6sin^2x-4sin^2x xxcos^2x=cos^2x#

#=>6(1-cos^2x)-4(1-cos^2x) cos^2x=cos^2x#

#=>6-6cos^2x-4cos^2x+4cos^4x=cos^2x#

#=>4cos^4x-11cos^2x+6=0#

#=>4cos^4x-8cos^2x-3cos^2x+6=0#

#=>4cos^2x(cos^2x-2)-3(cos^2x-2)=0#

#=(cos^2x-2)(4cos^2x-3)=0#

when #cos^2x-2=0=>cosxpmsqrt2#

but #-1<=cosx <=+1#
so this solution is not acceptable.

when

#(4cos^2x-3)=0#

#cosx=pmsqrt3/2#

For

#cosx=+sqrt3/2=cos(pi/6)#

#=>x=2npipmpi/6" where " n in ZZ#

For

#cosx=-sqrt3/2=-cos(pi/6)=cos((5pi)/6)#

#=>x=2npipm(5pi)/6" where " n in ZZ#