Answer:
The possible values of the angle \alpha that satisfy the given equation are the angles in the range [0, 60] degrees or the angles in the range [300, 360] degrees. This is because we showed that the value of cos(\alpha) must be less than or equal to 1/\sqrt{3} in order for the equation to hold.
Explanation:
It looks like you have entered some LaTeX code in your question, but it is not properly formatted. Here is the correct way to enter the equation:
cos(\alpha) + \frac{1}{\sqrt{3}} \cdot sin(\alpha) = 1
To solve this equation, we can use the identity cos^2(\alpha) + sin^2(\alpha) = 1. We can rewrite the left-hand side of the equation as follows:
cos^2(\alpha) + \frac{1}{3} \cdot sin^2(\alpha) = 1
Then, we can solve for sin^2(\alpha):
sin^2(\alpha) = 3 \cdot (1 - cos^2(\alpha))
Since sin^2(\alpha) is nonnegative, we know that cos^2(\alpha) must be less than or equal to 1/3. This means that cos(\alpha) must be less than or equal to 1/\sqrt{3}. Therefore, the possible values of \alpha that satisfy the given equation are the angles in the range [0, 60] degrees or the angles in the range [300, 360] degrees.
Note that this solution assumes that the angle \alpha is measured in degrees. If the angle is measured in radians, then the range of possible values for \alpha will be different.