Question

9cos^2x+5sinx=3sin^2x solve for x?

Answer 1

`x=nxx180^@-(-1)^n43.03^@`

Explanation:

`9cos^2x+5sinx=3sin^2x` can be written as

`9(1-sin^2x)+5sinx=3sin^2x`

or `9-9sin^2x+5sinx=3sin^2x`

or `12sin^2x-5sinx-9=0`

or `sinx=(5+-sqrt(5^2+4*12*9))/24`

= `(5+-sqrt457)/24`

= `(5+-21.3776)/24`

But `5+21.3776>24`, hence this leads to `sinx>1` and is not possible.

Hence `sinx=(5-21.3776)/24=-0.6824=sin(-43.03^@)`

and `x=nxx180^@-(-1)^n43.03^@`