Question

How do you solve `5sin^2x + 3cos^2x = 4`?

Answer 1

The solutions are `x=pi/4 ,x=(3pi)/4,x=(5pi)/4,x=(7pi)/4`

Explanation:

We know that `sin^2x+cos^2x=1` hence `cos^2x=1-sin^2x`

thus

#5sin^2x+3(1-sin^2x)=4=>5sin^2x+3-3sin^2x=4=> 2sin^2x=1=>sinx=+-sqrt2/2=>x=pi/4 ,x=(3pi)/4,x=(5pi)/4,x=(7pi)/4#