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.
Solve the equation for
x if 0≤x<2π.
4sin^2x+4cosx−5=0
θ=
3 answers
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
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.
2 answers