Solve 2cos2x + 4cosx - 1 = 0 for 0° ≤ x < 180°.{77°}{103°}Ø{73°}

To solve the equation 2cos2x + 4cosx - 1 = 0 for 0° ≤ x < 180°, we can use the substitution method.

Let's substitute cos2x with 1 - 2sin^2x:

2(1 - 2sin^2x) + 4cosx - 1 = 0

Expanding and rearranging the equation:

2 - 4sin^2x + 4cosx - 1 = 0

-4sin^2x + 4cosx + 1 = 0

Now, let's use the identity sin^2x + cos^2x = 1 to rewrite the equation:

-4(1 - cos^2x) + 4cosx + 1 = 0

-4 + 4cos^2x + 4cosx + 1 = 0

4cos^2x + 4cosx - 3 = 0

Now, we can solve this quadratic equation for cosx. Let's factorize it:

(2cosx - 1)(2cosx + 3) = 0

Setting each factor equal to zero:

2cosx - 1 = 0 or 2cosx + 3 = 0

Solving the first equation:

2cosx = 1

cosx = 1/2

x = arccos(1/2)

Using the unit circle or a calculator, we find two solutions for x:

x = 60° or x = 300°

Now, let's solve the second equation:

2cosx = -3

cosx = -3/2

x = arccos(-3/2)

Using the unit circle or a calculator, we find two solutions for x:

x = 120° or x = 240°

Therefore, the solutions for the equation 2cos2x + 4cosx - 1 = 0 for 0° ≤ x < 180° are:

x = 60°, 120°, 240°, 300°.