Solve the equation, where 0° ≤ x < 360°. Round approximate solutions to the nearest tenth of a degree. (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.)

3 sin x − 3 cos x = 1

3sinx = 1 + 3cosx

square both sides
9sin^2 x = 1 + 6cosx + 9cos^2 x
9(1 - cos^2 x) = 1 + 6cosx + 9cos^2 x
9 - 9cos^2 x = 1 + 6x + 9cos^2 x
18cos^2 x + 6x - 8 = 0
9cos^2 x + 3x - 4 = 0
cosx = ( -3 ± √153)/18
= -.85385 or .5205

if cosx = -.85385
x = 148.6° or x = 211.4°

if cosx = .5205
x = 58.6° or x = 301.4°

BUT , since we squared our equation, ALL solutions must be verified in the original equation.
Using my calculator, only x = 58.6° and x = 211.4° actually worked in the original equation.

To solve the equation 3 sin x − 3 cos x = 1, we can use the trigonometric identity cos x = sin(90°-x).

Substituting this into the equation, we have 3 sin x − 3 sin(90°-x) = 1.

Expanding the equation, we get 3 sin x − 3 sin 90° cos x + 3 cos 90° sin x = 1.

Simplifying further, we have 3 sin x − 3 cos x = 1.

Now, let's rearrange the equation by moving all the terms to one side: 3 sin x − 3 cos x - 1 = 0.

Next, we can use the trigonometric identity sin x = cos(90°-x) to substitute for sin x in the equation.

We have 3 cos(90°-x) − 3 cos x - 1 = 0.

Expanding this equation, we get 3 cos 90° cos x + 3 sin 90° sin x − 3 cos x - 1 = 0.

Simplifying further, we have 3 cos x + 3 sin x − 3 cos x - 1 = 0.

Canceling out the common terms, we are left with 3 sin x - 1 = 0.

To solve for sin x, divide both sides of the equation by 3: sin x = 1/3.

Now, we need to find the values of x between 0° and 360° where sin x is equal to 1/3.

Using a calculator or a trigonometric table, we can find that the two angles where sin x = 1/3 are approximately 19.5° and 160.5°.

Therefore, the solutions to the equation 3 sin x − 3 cos x = 1, where 0° ≤ x < 360°, are approximately 19.5° and 160.5°.

So the solution is x ≈ 19.5°, 160.5°.

To solve the equation 3 sin x − 3 cos x = 1, we can rearrange the equation and try to find the values of x that satisfy the equation.

Let's start by manipulating the equation:

3 sin x − 3 cos x = 1

Divide the entire equation by 3:

sin x − cos x = 1/3

Now, we can use the identity cos^2 x + sin^2 x = 1 to express sin x in terms of cos x:

sin^2 x = 1 - cos^2 x

Substitute this into the equation:

1 - cos^2 x - cos x = 1/3

Rearrange the equation:

cos^2 x + cos x - 2/3 = 0

Now, we have a quadratic equation in terms of cos x. We can solve this quadratic equation using the quadratic formula:

cos x = [-b ± √(b^2 - 4ac)] / (2a)

In this case, a = 1, b = 1, and c = -2/3.

cos x = [-1 ± √(1^2 - 4(1)(-2/3))] / (2(1))

cos x = [-1 ± √(1 + 8/3)] / 2

cos x = [-1 ± √(11/3)] / 2

Now, we can find the values of cos x by evaluating this expression.

cos x = (-1 + √(11/3)) / 2

cos x ≈ 0.551

cos x = (-1 - √(11/3)) / 2

cos x ≈ -1.218

Now, we need to find the corresponding values of sin x for each of these cos x values.

sin x = ± √(1 - cos^2 x)

For cos x ≈ 0.551, sin x ≈ ± √(1 - (0.551)^2) ≈ ± 0.835

For cos x ≈ -1.218, sin x ≈ ± √(1 - (-1.218)^2) ≈ ± 0.728

Now, we can find the corresponding values of x by using the inverse trigonometric functions.

For cos x ≈ 0.551:
x ≈ arcsin(0.835) ≈ 57.8°
x ≈ arccos(0.835) ≈ 32.2°

For cos x ≈ -1.218:
x ≈ arcsin(0.728) ≈ 46.5°
x ≈ arccos(0.728) ≈ 313.5°

Therefore, the approximate solutions to the equation 3 sin x − 3 cos x = 1 are:
x ≈ 57.8°, 32.2°, 46.5°, 313.5° (rounded to the nearest tenth of a degree).