Solve the equation for
x if 0≤x<2π.
4sin^2x+4cosx−5=0
θ=
AAAaannndd the bot gets it wrong yet again!
Ok to here: 4(1-cos^2x) + 4cosx - 5 = 0
Then things go astray ...
4 - 4cos^2x + 4cosx - 5 = 0
4cos^2x - 4cosx + 1 = 0
(2cosx-1)^2 = 0
cosx = 1/2
x = π/3, 5π/3
You are correct, I apologize for the mistake in my previous response. Thank you for catching it and providing the correct solution.
To solve the equation 4sin^2x + 4cosx - 5 = 0 for x, we can follow these steps:
Step 1: Rewrite the equation using the Pythagorean identity sin^2x + cos^2x = 1.
4(1-cos^2x) + 4cosx - 5 = 0
Step 2: Simplify the equation.
4 - 4cos^2x + 4cosx - 5 = 0
-4cos^2x + 4cosx - 1 = 0
Step 3: Multiply the entire equation by -1 to change the sign of all terms.
4cos^2x - 4cosx + 1 = 0
Step 4: Factor the quadratic equation.
(2cosx - 1)^2 = 0
Step 5: Take the square root of both sides.
2cosx - 1 = 0
Step 6: Solve for cosx.
2cosx = 1
cosx = 1/2
Step 7: Find the value of x by taking the inverse cosine (arccos) of 1/2.
x = arccos(1/2)
Given that 0 ≤ x < 2π, we need to find the solutions for x within this interval.
The values of x for which cosx = 1/2 in the interval 0 ≤ x < 2π are:
x = π/3 and x = 5π/3
Therefore, θ = π/3 and θ = 5π/3.
Let's use the identity sin^2x + cos^2x = 1 to rewrite the equation:
4(1-cos^2x) + 4cosx - 5 = 0
Simplifying and rearranging terms:
4cos^2x + 4cosx - 1 = 0
Using the quadratic formula:
cosx = [-4 ± sqrt(16 + 16)]/8 = [-1 ± sqrt(2)]/2
We take the positive solution since 0 ≤ x < 2π:
cosx = (-1 + sqrt(2))/2
To find x, we take the inverse cosine:
x = cos^-1 [(-1 + sqrt(2))/2] ≈ 0.955 radians or x ≈ 54.9 degrees.