Solve for x on the interval [0, 2pi] 1. tan^2x - 1 = 0 2. cos^2x + 3cosx + 2 = 0 3. 2sin^2x - sinx = 1?

Answer 1

It would be easier to read your question if you put separate problems on separate lines.

1. tan^2 x - 1 = 0
<=> (tan x + 1) (tan x - 1) = 0
<=> tan x = ±1
<=> x = π/4, 3π/4, 5π/4, 7π/4.

2. cos^2 x + 3 cos x + 2 = 0
<=> (cos x + 2) (cos x + 1) = 0
<=> cos x = -2 or -1
<=> cos x = -1 since cos x = -2 is impossible
<=> x = π.

3. 2 sin^2 x - sin x = 1
<=> 2 sin^2 x - sin x - 1 = 0
<=> (2 sin x + 1) (sin x - 1) = 0
<=> sin x = -1/2 or 1
<=> x = π/2, 7π/6, 11π/6.