What value(s) of theta (0 less than or equal to theta less than or equal to 2pi) solve the following equation: cos squared theta - cos theta - 6 = 0

cos^2 Ø - sin^2 Ø = 6

cos 2Ø = 6

no solution, the cosine of anything cannot be greater than 1
do you have a typo?

No, I don't just my professor trying to trick us, thanks for the help!

To solve the equation cos²θ - cosθ - 6 = 0, we can rewrite it as a quadratic equation by letting x = cosθ. Therefore:

x² - x - 6 = 0

Now, we can factor the quadratic equation:

(x - 3)(x + 2) = 0

Setting each factor to zero, we get:

x - 3 = 0 OR x + 2 = 0

Solving for x, we have:

x = 3 OR x = -2

Since x represents cosθ, we need to find the values of θ that correspond to these x values. To do so, we can make use of the inverse cosine function (also known as arccosine or cos⁻¹).

When x = 3, we have:

θ = cos⁻¹(3)

However, the cosine function only outputs values between -1 and 1, so cos⁻¹(3) is not defined. Therefore, there are no solutions in the range 0 ≤ θ ≤ 2π.

When x = -2, we have:

θ = cos⁻¹(-2)

The cos⁻¹(-2) is defined, but we need to find the principal value of the inverse cosine function, which lies in the range 0 ≤ θ ≤ π. In this range, cosθ = -2 does not have a real solution, so there are no solutions in this interval as well.

Therefore, the equation cos²θ - cosθ - 6 = 0 has no solutions in the interval 0 ≤ θ ≤ 2π.