2sin^2(x) = 2 + cos(x)
interval [0,2pi)
How do I solve this? Help would be greatly appreciated!
2(1-cos^2 x) = 2 + cos x
2 - 2 cos^2 x = 2 + cos x
2 cos^2 x + cos x = 0
cos x (2 cos x + 1 ) = 0
cos x = 0 when x = pi/2 or 90 degrees or 3 pi/2 or 270 deg
or
cos x = -1/2 at x = 2pi/3 or 120 degrees or at 4 pi/3 or 240 degrees
To solve the equation 2sin^2(x) = 2 + cos(x) over the interval [0, 2π), you can follow these steps:
Step 1: Rearrange the equation
Start by rewriting the equation in terms of trigonometric identities. Notice that sin^2(x) = 1 - cos^2(x), so we can substitute this into our equation:
2(1 - cos^2(x)) = 2 + cos(x)
Step 2: Simplify the equation
Distribute 2 on the left side:
2 - 2cos^2(x) = 2 + cos(x)
Step 3: Move all terms to one side
Subtract 2 and cos(x) from both sides of the equation:
-2cos^2(x) - cos(x) = 0
Step 4: Factor out common terms
Factor out cos(x) from the left side of the equation:
cos(x)(-2cos(x) - 1) = 0
Step 5: Set each factor equal to zero
Now you have two separate equations to solve:
cos(x) = 0
-2cos(x) - 1 = 0
Step 6: Solve each equation
For cos(x) = 0, you can find the solutions by using the unit circle or by knowing that the cosine is equal to zero at π/2 and 3π/2.
For -2cos(x) - 1 = 0, solve for cos(x) by isolating the variable:
-2cos(x) = 1
cos(x) = -1/2
You can find the solutions for cos(x) = -1/2 by using the unit circle or knowing that the cosine is equal to -1/2 at 2π/3 and 4π/3.
Step 7: Determine the solutions within the given interval
In the interval [0, 2π), the solutions for cos(x) = 0 are π/2 and 3π/2.
The solutions for cos(x) = -1/2 are 2π/3 and 4π/3.
So, the solutions to the original equation over the interval [0, 2π) are:
x = π/2, 3π/2, 2π/3, 4π/3