How to solve this trigonometric equation?

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1 Answer
Apr 1, 2018

Answer is (3).

Explanation:

5(tan^2x-cos^2x)=2cos2x+95(tan2xcos2x)=2cos2x+9

hArr5(sec^2x-1)-5cos^2x=2(2cos^2x-1)+95(sec2x1)5cos2x=2(2cos2x1)+9

or 5sec^2x-9cos^2x-12=05sec2x9cos2x12=0 and multiplying by cos^2xcos2x we get

5-9cos^4x-12cos^2x=059cos4x12cos2x=0

i.e. 9cos^4x+12cos^2x-5=09cos4x+12cos2x5=0

and cos^2x=(-12+-sqrt(144+180))/18cos2x=12±144+18018

= (-12+-18)/1812±1818

i.e. cos^2x=1/3cos2x=13 as we cannot have cos^2xcos2x as negative

Hence cos2x=2cos^2x-1=2*1/3-1=-1/3cos2x=2cos2x1=2131=13

and cos4x=2cos^2 2x-1=2(-1/3)^2-1=-7/9cos4x=2cos22x1=2(13)21=79

Hence, answer is (3).