Solve for x in the interval [0,2pi) sin^2x+2cosx=2
sin^2x+2cosx=2
1 - cos^2x + 2cosx - 2 = 0
cos^2x - 2cosx + 1 = 0
(cosx - 1)^2 = 0
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cosx = 1
Can you take it from there?
To solve the equation sin^2x + 2cosx = 2 on the interval [0,2π), we can use some trigonometric identities to simplify the equation.
Step 1: Rewrite the equation using the identity sin^2x = 1 - cos^2x.
1 - cos^2x + 2cosx = 2
Step 2: Rearrange the equation to get a quadratic equation in terms of cos(x).
cos^2x + 2cosx - 1 = 0
Step 3: Solve the quadratic equation using the quadratic formula.
The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by x = (-b ± √(b^2 - 4ac)) / (2a).
In our case, a = 1, b = 2, and c = -1. Plugging these values into the quadratic formula, we get:
cos(x) = [-(2) ± √((2)^2 - 4(1)(-1))] / (2(1))
cos(x) = [-2 ± √(4 + 4)] / 2
cos(x) = [-2 ± √(8)] / 2
cos(x) = [-2 ± 2√2] / 2
cos(x) = -1 ± √2
Step 4: Simplify the solutions and find the corresponding values for x.
Since cos(x) = -1 ± √2, we need to find the values of x for which cos(x) = -1 + √2 and cos(x) = -1 - √2.
For cos(x) = -1 + √2, we can use the inverse cosine (arccos) function to find the corresponding value of x.
x = arccos(-1 + √2)
For cos(x) = -1 - √2, we can use the inverse cosine (arccos) function to find the corresponding value of x.
x = arccos(-1 - √2)
Step 5: Calculate the values of x within the specified interval.
Since the given interval is [0,2π), substitute the values of x obtained from step 4 and find the values of x within this interval.
x = arccos(-1 + √2) is approximately 0.245 rad (or approximately 14.04 degrees).
x = arccos(-1 - √2) is approximately 2.897 rad (or approximately 165.96 degrees).
Therefore, the solutions for x in the interval [0,2π) are x ≈ 0.245 and x ≈ 2.897.