Solve the equation for exact solutions in the interval 0 ≤ x < 2π. (Enter your answers as a comma-separated list.)

cos 2x cos x − sin 2x sin x = 0

(cos^2 x - sin^2 x)(cosx) - (2sinxcosx)(sinx) = 0

divide by cosx
cos^2 x - sin^2 x - 2sin^2 x = 0
cos^2 x = 3sin^2 x
sin^2 x/cos^2 x = 1/3
tan^2 x = 1/3
tanx = ± 1/√3
so x could in in any of the 4 quadrants

I know tan 30° = 1/√3
and 30° = π/6 radians

so x = 30° , 150° , 210° , 330°
or
x = π/6 , 5π/6 , 7π/6 , 11π/6

To solve the equation cos(2x)cos(x) - sin(2x)sin(x) = 0, we can simplify it using trigonometric identities.

Using the identity cos(A - B) = cos(A)cos(B) + sin(A)sin(B), we can rewrite the equation as:

cos(2x - x) = 0

Simplifying further, we have:

cos(x) = 0

Now, let's find the solutions for cos(x) = 0 in the interval 0 ≤ x < 2π. To do this, we need to find the angles where the cosine function equals zero.

The cosine function equals zero at two points within one period (2π). These angles are:

x = π/2 and x = 3π/2

Therefore, the exact solutions in the interval 0 ≤ x < 2π are:

x = π/2, 3π/2

To solve the equation cos 2x cos x - sin 2x sin x = 0, we can use the trigonometric identity for the cosine of the difference of angles:

cos(A - B) = cos A cos B + sin A sin B

Comparing this identity to the given equation, we can see that:

cos 2x cos x - sin 2x sin x = cos (2x - x) = cos x

So, the equation is simplified to:

cos x = 0

To find the exact solutions in the interval 0 ≤ x < 2π, we need to find the values of x where the cosine function equals zero.

In the interval 0 ≤ x < 2π, the cosine function equals zero at x = π/2 and x = 3π/2.

Therefore, the exact solutions in the interval 0 ≤ x < 2π are x = π/2 and x = 3π/2.

So, the answer is x = π/2, 3π/2.