1. ## Multiple Solutions

find all the solutions from 0 to 2pi

2cos^2x = 13sinx-5

2. Originally Posted by jumpman23
find all the solutions from 0 to 2pi

2cos^2x = 13sinx-5
$\cos^2{x} = 1 - \sin^2{x}$ ... solve the resulting quadratic for $\sin{x}$

btw ... why are these basic trig problems on the calculus page?