smallest degree that satisfies 2cos^2x-3cosx=2
2cos^2x-3cosx=2
2cos^2x - 3cosx - 2 = 0
(2cosx + 1)(cosx - 2) = 0
cosx = -1/2 or cosx = 2, the last part is not possible
cosx = -1/2, so the smallest angle is in quadrant II
x = 120° or 2π/3 rads , assuming you wanted the smallest positive angle
To find the smallest degree that satisfies the equation 2cos^2x - 3cosx = 2, we can follow these steps:
Step 1: Rewrite the equation in a quadratic form.
2cos^2x - 3cosx - 2 = 0
Step 2: Let's use a substitution to simplify the equation. We substitute cos(x) with a variable, let's say t.
2t^2 - 3t - 2 = 0
Step 3: Solve the quadratic equation by factoring, completing the square, or using the quadratic formula.
In this case, factoring might not be straightforward, so let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 2, b = -3, and c = -2.
t = (-(-3) ± √((-3)^2 - 4(2)(-2))) / (2(2))
t = (3 ± √(9 + 16)) / 4
t = (3 ± √25) / 4
t = (3 ± 5) / 4
Step 4: Solve for t in both cases (positive and negative square root).
Case 1: (3 + 5) / 4
t1 = (8) / 4
t1 = 2
Case 2: (3 - 5) / 4
t2 = (-2) / 4
t2 = -0.5
Step 5: Substitute back in the original variable cos(x).
Case 1: cos(x) = 2
Since the cosine function's range is limited to -1 ≤ cos(x) ≤ 1, there is no solution for cos(x) = 2. Therefore, this case is invalid.
Case 2: cos(x) = -0.5
To find the smallest degree that satisfies this equation, we can use the inverse cosine function (also known as arccos or cos^-1).
cos^-1(-0.5) ≈ 120°
Therefore, the smallest degree that satisfies the equation 2cos^2x - 3cosx = 2 is approximately 120°.