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

The solutions are $x = \frac{\pi}{4} , x = \frac{3 \pi}{4} , x = \frac{5 \pi}{4} , x = \frac{7 \pi}{4}$
We know that ${\sin}^{2} x + {\cos}^{2} x = 1$ hence ${\cos}^{2} x = 1 - {\sin}^{2} x$
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