Question

Solve cos x - square 1-3 cos^2 x = 0 given that 0° < x < 360°

Answer 1

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.