Solve 2cosx-sin^x +2 =0 for all real values of x
To solve the equation 2cos(x) - sin^2(x) + 2 = 0 for all real values of x, we'll go through the following steps:
Step 1: Rewrite the equation using trigonometric identities.
Let's use the identity sin^2(x) = 1 - cos^2(x) to rewrite the equation:
2cos(x) - (1 - cos^2(x)) + 2 = 0
Simplifying, we get:
2cos(x) - 1 + cos^2(x) + 2 = 0
Rearranging terms, we have:
cos^2(x) + 2cos(x) + 1 = 0
Step 2: Factor the quadratic equation.
The equation can be factored into:
(cos(x) + 1)^2 = 0
Step 3: Solve for cos(x).
Setting (cos(x) + 1)^2 = 0, we have two cases to consider:
Case 1: cos(x) + 1 = 0
Solving this equation, we find:
cos(x) = -1
Case 2: (cos(x) + 1)^2 = 0
Taking the square root of both sides, we get:
cos(x) + 1 = 0
cos(x) = -1
Step 4: Find the values of x.
The cosine function is equal to -1 at π radians (180 degrees) or an odd multiple of π radians (or degrees).
Therefore, the solution for all real values of x is:
x = π + 2πk (where k is an integer)
In conclusion, the equation 2cos(x) - sin^2(x) + 2 = 0 is satisfied for all real values of x at x = π + 2πk (where k is an integer).
You must have a typo.
I will assume you meant ...
2cosx-sin^2 x +2 =0
2cosx - (1 - cos^2x) + 2 = 0
cos^2x + 2cosx + 1 = 0
(cosx + 1)^2 = 0
cosx + 1 = 0
cosx = -1
x = 180º or x = pi radians