Question

How to solve this trigonometric equation?

Answer 1

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).