Find all solutions of the equation in the interval [0, 2π).
cos(x + 9π) − cos x + 1 = 0
9/2 =4 1/2
so 4 times all the way + one time halfway around which is pi
so you really have cos (x+pi) - cos x + 1 = 0
but cos (a+b) = cos a cos b - sin a sinb
so it is
cos x * (-1) - sin x * (0) - cos x + 1 = 0
-2 cos x = -1
cos x = 0.5
60 and 300 deg
To find all the solutions of the given equation in the interval [0, 2π), we can use trigonometric identities and algebraic manipulations. Let's break down the steps:
1. Use the sum-to-product identity for cosine:
cos(A) - cos(B) = -2 * sin((A + B) / 2) * sin((A - B) / 2)
2. Apply the identity to the given equation:
-2 * sin((x + 9π + x) / 2) * sin((x + 9π - x) / 2) + 1 = 0
Simplifying further:
-2 * sin((2x + 9π) / 2) * sin((9π) / 2) + 1 = 0
3. Use the double-angle identity for sine:
sin(2θ) = 2 * sin(θ) * cos(θ)
Applying this to the equation:
-2 * sin(x + 9π) * sin(9π / 2) + 1 = 0
4. Expand sin(9π) and sin(9π / 2) using periodicity:
sin(9π) = sin(2π + 7π) = sin(7π)
sin(9π / 2) = sin(π + 4π / 2) = sin(π + 2π) = sin(3π)
The equation becomes:
-2 * sin(x + 7π) * sin(3π) + 1 = 0
5. Replace sin(7π) and sin(3π) with their respective values (-1):
-2 * sin(x + 7π) * (-1) + 1 = 0
6. Simplify the equation further:
2 * sin(x + 7π) + 1 = 0
7. Solve for sin(x + 7π):
2 * sin(x + 7π) = -1
sin(x + 7π) = -1/2
8. We know that sinθ = -1/2 has solutions at π/6 and 5π/6. Therefore, we have:
x + 7π = π/6 or x + 7π = 5π/6
9. Solve for x:
For the first equation,
x = π/6 - 7π
x = -43π/6
For the second equation,
x = 5π/6 - 7π
x = -37π/6
10. Since we are looking for solutions in the interval [0, 2π), we need to adjust the solutions. Adding 2π to each solution, we get:
x = -43π/6 + 12π/6 = -31π/6
x = -37π/6 + 12π/6 = -25π/6
11. Convert the solutions to the interval [0, 2π):
x = | -31π/6 | mod 2π
x = 11π/6
x = | -25π/6 | mod 2π
x = 7π/6
Therefore, the solutions in the interval [0, 2π) are x = 11π/6 and x = 7π/6.