Give a list of solutions, in the interval [0, 360°), in degrees, of the equation 2cos^24θ + 5 cos 4θ = 3.
your equation is
(2cos4θ - 1)(cos4θ + 3) = 0
so, find all the angles θ where
cos4θ = 1/2 or cos4θ = -3
Note that cos4θ has period pi/2, so you will have more solutions than normal, since the domain covers 4 periods.
To find the solutions in the interval [0, 360°), we'll start by simplifying the equation:
2cos^2(4θ) + 5cos(4θ) = 3
Let's make a substitution to simplify the equation. Replace cos(4θ) with x:
2x^2 + 5x = 3
Rearrange the equation to form a quadratic equation:
2x^2 + 5x - 3 = 0
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = 2, b = 5, and c = -3. Substituting these values into the quadratic formula, we have:
x = (-5 ± √(5^2 - 4 * 2 * -3)) / (2 * 2)
Simplifying:
x = (-5 ± √(25 + 24)) / 4
x = (-5 ± √49) / 4
x = (-5 ± 7) / 4
So, we have two possible values for x:
x1 = (-5 + 7) / 4 = 2 / 4 = 0.5
x2 = (-5 - 7) / 4 = -12 / 4 = -3
Now, we'll find the values of 4θ that correspond to these values of x. Recall that x = cos(4θ).
For x = 0.5:
cos(4θ) = 0.5
Let's find the angles that give a cosine value of 0.5 in the interval [0, 360°).
θ1 = cos^-1(0.5) ≈ 60°
θ2 = 360° - θ1 ≈ 300°
For x = -3:
cos(4θ) = -3
Let's find the angles that give a cosine value of -3. However, cosine values are always between -1 and 1, so there are no solutions in the interval [0, 360°).
Therefore, the solutions in the interval [0, 360°), in degrees, of the equation 2cos^24θ + 5 cos 4θ = 3 are:
θ1 ≈ 60°
θ2 ≈ 300°