2〖sen〗^2 x+3 senx+1=0 2cotxsecx+ 2secx+cotx+1=0

senx〖cos〗^2 x=senx

2〖cos〗^2 x+2senx-12=0

2〖csc〗^2 x+〖cot〗^2 x-3=0

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

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step-by-step explanation: