Rearrange the equation as
cos(x) - √(1 - 3 cos²x) = 0
cos(x) = √(1 - 3 cos²x)
Note that both sides have to be positive, since the square root function is non-negative. If we end up with any solutions that make cos(x) < 0, we must throw them out.
Take the square of both sides.
(cos(x))² = (√(1 - 3 cos²(x))²
cos²(x) = 1 - 3 cos²(x)
4 cos²(x) = 1
cos²(x) = 1/4
cos(x) = ± √(1/4)
cos(x) = ± 1/2
We throw out the negative case and we're left with
cos(x) = 1/2
This has general solution
x = arccos(1/2) + 360° n or x = -arccos(1/2) + 360° n
(where n is any integer)
x = 60° + 360° n or x = -60° + 360° n
Now just pick out the solutions that fall in the interval 0° ≤ x < 360°.
• From the first solution set, we have x = 60° when n = 0.
• From the second set, we have x = 300° when n = 1.
So the answer is D.
