2 sin²(x) + 3 sin(x) + 1 = 0
(2 sin(x) + 1) (sin(x) + 1) = 0
2 sin(x) + 1 = 0 OR sin(x) + 1 = 0
sin(x) = -1/2 OR sin(x) = -1
The first equation gives two solution sets,
x = sin⁻¹(-1/2) + 2nπ = -π/6 + 2nπ
x = π - sin⁻¹(-1/2) + 2nπ = 5π/6 + 2nπ
(where n is any integer), while the second equation gives
x = sin⁻¹(-1) + 2nπ = -π/2 + 2nπ
2 cot(x) sec(x) + 2 sec(x) + cot(x) + 1 = 0
2 sec(x) (cot(x) + 1) + cot(x) + 1 = 0
(2 sec(x) + 1) (cot(x) + 1) = 0
2 sec(x) + 1 = 0 OR cot(x) + 1 = 0
sec(x) = -1/2 OR cot(x) = -1
cos(x) = -2 OR tan(x) = -1
The first equation has no (real) solutions, since -1 ≤ cos(x) ≤ 1 for all (real) x. The second equation gives
x = tan⁻¹(-1) + nπ = -π/4 + nπ
sin(x) cos²(x) = sin(x)
sin(x) cos²(x) - sin(x) = 0
sin(x) (cos²(x) - 1) = 0
sin(x) (-sin²(x)) = 0
sin³(x) = 0
sin(x) = 0
x = sin⁻¹(0) + 2nπ = 2nπ
2 cos²(x) + 2 sin(x) - 12 = 0
2 (1 - sin²(x)) + 2 sin(x) - 12 = 0
-2 sin²(x) + 2 sin(x) - 10 = 0
sin²(x) - sin(x) + 5 = 0
Using the quadratic formula, we get
sin(x) = (1 ± √(1 - 20)) / 2 = (1 ± √(-19)) / 2
but the square root contains a negative number, which means there is no real solution.
2 csc²(x) + cot²(x) - 3 = 0
2 (cot²(x) + 1) + cot²(x) - 3 = 0
3 cot²(x) - 1 = 0
cot²(x) = 1/3
tan²(x) = 3
tan(x) = ± √3
x = tan⁻¹(√3) + nπ OR x = tan⁻¹(-√3) + nπ
x = π/3 + nπ OR x = -π/3 + nπ