Solve the equation for x if 0 ≤ x < 2𝜋. Give your answer in radians using exact values only. (Enter your answers as a comma-separated list.)2 cos2 x − 3 sin x − 3 = 0x =

To solve the given trigonometric equation, we first need to express it in terms of one trigonometric function. The equation is 2 cos²x - 3 sinx - 3 = 0.

We know that sin²x + cos²x = 1, so we can express cos²x as 1 - sin²x. Substituting this into the equation gives us:

2(1 - sin²x) - 3sinx - 3 = 0 2 - 2sin²x - 3sinx - 3 = 0 2sin²x + 3sinx - 2 = 0

This is a quadratic equation in terms of sinx. We can solve it by using the quadratic formula:

sinx = [-b ± sqrt(b² - 4ac)] / 2a

Here, a = 2, b = 3, and c = -2. Substituting these values into the formula gives us:

sinx = [-3 ± sqrt((3)² - 42(-2))] / 2*2 sinx = [-3 ± sqrt(9 + 16)] / 4 sinx = [-3 ± sqrt(25)] / 4 sinx = [-3 ± 5] / 4

This gives us two possible solutions:

sinx = 2/4 = 0.5 sinx = -8/4 = -2

However, the range of sinx is -1 ≤ sinx ≤ 1, so sinx = -2 is not a valid solution.

Therefore, the only valid solution in terms of sinx is sinx = 0.5.

To find the corresponding values of x, we use the fact that sinx = 0.5 at x = π/6 and x = 5π/6 in the range 0 ≤ x < 2π.

So, the solutions to the equation are x = π/6, 5π/6.