Question

2(sinx)^2-5cosx-4=0
Solve for all solutions between [0,2pi]

Answer 1

change sin^2x to 1-cos^2x, then
2-2cos^2x-5cosx-4=0
let u= cosx
2-2u^2-5u-4=0
2u^2+5u+2=0
(2u+1)(U+2)=0
u=-1/2 u=-2
cosx=-1/2
x=-2PI/3, 2/3 PI
between o, 2PI then x= 2/3 PI, and 4/3 PI

Answer 2

put those two angles into the original equation, and check.