Solve the equation in the indicated domain.

1. 2cos^2 x-5cos x +2=0
x E[0,360)

2. sin^2 x+5 sin x +6=0
x E[0,360)

*assume x is theta

1. To solve the equation 2cos^2(x) - 5cos(x) + 2 = 0 in the domain x ∈ [0, 360), we can use the quadratic formula. Let's define cos(x) as t. The equation becomes:

2t^2 - 5t + 2 = 0

Applying the quadratic formula, which states that for an equation of the form at^2 + bt + c = 0, the solutions are given by:

t = (-b ± √(b^2 - 4ac)) / 2a

In our case, a = 2, b = -5, and c = 2. Substituting these values into the quadratic formula:

t = (-(-5) ± √((-5)^2 - 4(2)(2))) / (2(2))
t = (5 ± √(25 - 16)) / 4
t = (5 ± √9) / 4

Taking the positive solution first:

t = (5 + 3) / 4
t = 8 / 4
t = 2

This positive solution corresponds to cos(x) = 2. However, the range of cos(x) is [-1, 1], so this solution is not valid.

Now, taking the negative solution:

t = (5 - 3) / 4
t = 2 / 4
t = 0.5

This solution corresponds to cos(x) = 0.5. To find the values of x in the domain [0, 360) that satisfy this, we can use the inverse cosine function:

x = acos(0.5)

Using a calculator, we find x ≈ 60 or x ≈ 300. Thus, the solutions to the equation 2cos^2(x) - 5cos(x) + 2 = 0 in the domain [0, 360) are x ≈ 60 degrees and x ≈ 300 degrees.

2. To solve the equation sin^2(x) + 5sin(x) + 6 = 0 in the domain x ∈ [0, 360), we can use a similar approach. Let's define sin(x) as t. The equation becomes:

t^2 + 5t + 6 = 0

We can factorize this quadratic equation:

(t + 2)(t + 3) = 0

From this, we have two possibilities for t:

t + 2 = 0 --> t = -2
t + 3 = 0 --> t = -3

Now, let's substitute back sin(x) for t:

sin(x) = -2 --> x = asin(-2)
sin(x) = -3 --> x = asin(-3)

However, the range of sin(x) is [-1, 1]. Therefore, there are no solutions to the equation sin^2(x) + 5sin(x) + 6 = 0 in the domain [0, 360).

To solve the equations, we can use quadratic equations as both equations involve trigonometric functions squared.

1. 2cos²x - 5cosx + 2 = 0:

Let's solve this equation step by step:

Step 1: Substitute cosx with a variable, let's say u:
2u² - 5u + 2 = 0

Step 2: Factor the quadratic equation:
(2u - 1)(u - 2) = 0

Step 3: Set each factor equal to zero and solve for u:
2u - 1 = 0 -> u = 1/2
u - 2 = 0 -> u = 2

Step 4: Substitute back cosx for u:
cosx = 1/2 or cosx = 2

Step 5: Determine the values of x within the given domain [0, 360):
To find the angle values corresponding to cosx = 1/2 and cosx = 2, you can use an inverse cosine function (cos^(-1)) on a calculator or reference table.

For cosx = 1/2, we have two possible solutions in the given domain:
x₁ = cos^(-1)(1/2) = 60 degrees
x₂ = 360 - cos^(-1)(1/2) = 300 degrees

Since cosx cannot be equal to 2 within the range [0, 360), there is no solution for cosx = 2 in the given domain.

Therefore, the solutions for the equation 2cos²x - 5cosx + 2 = 0 in the domain [0, 360) are:
x = 60 degrees and x = 300 degrees.

2. sin²x + 5sinx + 6 = 0:

Again, we will solve this equation step by step:

Step 1: Substitute sinx with a variable, let's say v:
v² + 5v + 6 = 0

Step 2: Factor the quadratic equation:
(v + 2)(v + 3) = 0

Step 3: Set each factor equal to zero and solve for v:
v + 2 = 0 -> v = -2
v + 3 = 0 -> v = -3

Step 4: Substitute back sinx for v:
sinx = -2 or sinx = -3

Step 5: Determine the values of x within the given domain [0, 360):
To find the angle values corresponding to sinx = -2 and sinx = -3, you can use an inverse sine function (sin^(-1)) on a calculator or reference table.

However, there are no real solutions for sinx = -2 and sinx = -3 within the range of [-1, 1]. Therefore, the equation sin²x + 5sinx + 6 = 0 has no solution in the domain [0, 360).