Solve the following equations for x, if 0 ≤ x ≤ 2π.

i. 3csc² x – 4 = 0
ii. 4cos² x + 2cos x – 2 = 0

help me solve pls

(i) 3 csc²(x) - 4 = 0

3 csc²(x) = 4

csc²(x) = 4/3

sin²(x) = 3/4

sin(x) = ± √3/2

x = arcsin(√3/2) + 2nπ  or   x = arcsin(-√3/2) + 2nπ

x = π/3 + 2nπ   or   x = -π/3 + 2nπ

where n is any integer. The general result follows from the fact that sin(x) is 2π-periodic.

In the interval 0 ≤ x ≤ 2π, the first family of solutions gives x = π/3 and x = 4π/3 for n = 0 and n = 1, respectively; the second family gives x = 2π/3 and x = 5π/3 for n = 1 and n = 2.

(ii) 4 cos²(x) + 2 cos(x) - 2 = 0

2 cos²(x) + cos(x) - 1 = 0

(2 cos(x) - 1) (cos(x) + 1) = 0

2 cos(x) - 1 = 0   or   cos(x) + 1 = 0

2 cos(x) = 1   or   cos(x) = -1

cos(x) = 1/2   or   cos(x) = -1

[x = arccos(1/2) + 2nπ   or   x = 2π - arccos(1/2) + 2nπ]   or   x = arccos(-1) + 2nπ

[x = π/3 + 2nπ   or   x = 5π/3 + 2nπ]   or   x = π + 2nπ

For 0 ≤ x ≤ 2π, the solutions are x = π/3, x = 5π/3, and x = π.

  the correct choice is marked

step-by-step explanation:

the first two graphs show multiple turning points. the last graphs matches the description in every detail.