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+9

hArr5(sec^2x-1)-5cos^2x=2(2cos^2x-1)+9

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

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

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

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

= (-12+-18)/18

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

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

and cos4x=2cos^2 2x-1=2(-1/3)^2-1=-7/9

Hence, answer is (3).