How do i solve these trigonometric ? find all solutions in the interval [0, 2π). (sin x)(cos x) = 0 find all solutions in the interval [0, 2π). cos2x + 2 cos x + 1 = 0 find all solutions to the equation. cos2x + 2 cos x + 1 = 0 (i am not looking for easy answers to cheat, i just need on how to solve a bit confused and my teacher is not avail),

1) (sin x)(cos x) = 0 This means that either sin x = 0, or cos x = 0. Sin x = 0 when x = 0 or pi. cos x = 0 when x = pi/2 or 3pi/2. 2) cos2x + 2cosx + 1 = 0 First, expand cos2x. Simple trigonometric identity states that cos2x = (cos x)^2 - (sin x)^2.Now, substituting this equation, we get cos2x + 2cosx + 1 = (cos x)^2 - (sin x)^2 + 2cosx + 1. Note that we can now utilize the identity (cos x)^2 = 1 - (sin x)^2 to obtain (cos x)^2 + 2cosx + 1 - (sin x)^2 = 2(cos x)^2 + 2(cos x) = 0. Now, we can see that 2(cos x)^2 + 2(cos x) = 2(cos x)(cos x + 1) = 0. This is true when cos x = 0 or cos x + 1 = 0. Cos x = 0 when x = pi/2 or 3pi/2. cos x = -1 when x = pi. Therefore, x = pi/2, pi or 3pi/2.