Question

1. Solve for x in the interval [0 , 2 π ).
(a) 2sin^2x=sin x
(b) 3sin^2x+cos^2x=2
(c) 1+3cosx=cos2x
(d) sin2x+sin x=0

Answer 1

a) \(2\sin^2x-\sin x=0 \implies \sin x(2\sin{x}-1)=0 \implies \sin x=0 \text{ or } \sin{x}=\frac{1}{2}.\) You can proceed form here.

Answer 2

dont know hot to solve it