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

1 Answer
Oct 11, 2017

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