Find all degree solutions. (Enter your answers as a comma-separated list. Let k be any integer.)

2 cos2 6θ + 3 cos 6θ + 1 = 0

2 cos2 6θ + 3 cos 6θ + 1 = 0

(2cos6θ+1)(cos6θ+1) = 0
cos6θ = -1/2 or -1

so,
6θ = 2π/3 or 4π/3
θ = π/9+kπ/3 or 2π/9+kπ/3, k=0..5

6θ = π
θ = π+kπ/3, k=0..5

You add kπ/3 since the solutions are (6θ+2π)k

how would I convert these to degrees?

why would you want to?

Ok, if you must, recall that π = 180°

Thank you. The answer came out to

20degree + 60degreeK, 30degree + 60 degreeK, 40degree + 60degreek.

right, but I'd write it as

(20+60k)°, (30+60k)°, (40+60k)°

I see you caught my typo.

Solve the equation for all degree solutions and if

0° ≤ θ < 360°.
Do not use a calculator. (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.)
2 sin θ −

3
= 0

(a) all degree solutions (Let k be any integer.)
θ =


(b) 0° ≤ θ < 360°
θ =

To find all the degree solutions for the equation 2cos^2 6θ + 3cos 6θ + 1 = 0, we can start by making a substitution.

Let's substitute x = cos 6θ. Now, we have a quadratic equation in terms of x:

2x^2 + 3x + 1 = 0

To solve this quadratic equation, we can use the quadratic formula. The quadratic formula states that for an equation in the form ax^2 + bx + c = 0, the solutions for x can be found using the formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

Applying this formula to our quadratic equation, we have:

x = (-(3) ± √((3^2) - 4(2)(1))) / (2(2))
x = (-3 ± √(9 - 8)) / 4
x = (-3 ± √1) / 4

Simplifying further, we get:

x = (-3 ± 1) / 4

We have two possible solutions for x:

1) x = (-3 + 1) / 4 = -2/4 = -1/2
2) x = (-3 - 1) / 4 = -4/4 = -1

Now, we need to substitute back x = cos 6θ to find the solutions for θ.

For x = -1/2:
cos 6θ = -1/2

One possible solution is when 6θ = 120 degrees:
6θ = 120
θ = 120/6
θ = 20 degrees

Another possible solution is when 6θ = 240 degrees:
6θ = 240
θ = 240/6
θ = 40 degrees

For x = -1:
cos 6θ = -1

One possible solution is when 6θ = 180 degrees:
6θ = 180
θ = 180/6
θ = 30 degrees

Therefore, the degree solutions for the equation 2cos^2 6θ + 3cos 6θ + 1 = 0 are 20 degrees, 30 degrees, and 40 degrees.

In a comma-separated list, the solutions are: 20, 30, 40.