Find the exact solutions of the equation in the interval [0,2pi).
sin(x/2)+cos(x)=0
recall the half-angel conversions
cos 2A = 1 - 2sin^2 A
so
cos x = = 1 - 2sin^2 (x/2)
sin(x/2) + 1 - 2sin^2 (x/2) = 0
2sin^2(x/2) - sin(x/2) - 1 = 0
(2sin(x/2) + 1)(sin(x/2) - 1) = 0
sin(x/2) = -1/2 or sin(x/2) = 1
x/2 = 7π/6 or x/2 = 11π/6 or x/2 = π/2
x = 7π/3 or x = 11π/3 or x = π (420° , 660· , 180° )
so for the given domain
x = π
To find the exact solutions of the equation sin(x/2) + cos(x) = 0 in the interval [0, 2π), follow these steps:
1. Start by isolating either sin(x/2) or cos(x) in terms of the other trigonometric function.
Let's isolate sin(x/2):
sin(x/2) = -cos(x)
2. Square both sides of the equation to eliminate the square root.
(sin(x/2))^2 = (-cos(x))^2
sin^2(x/2) = cos^2(x)
3. Apply the Pythagorean identity sin^2(x) + cos^2(x) = 1.
sin^2(x/2) = 1 - sin^2(x/2)
2sin^2(x/2) - 1 = 0
4. Now, apply the double-angle formula for sine: sin(2θ) = 2sin(θ)cos(θ)
Using the double-angle formula for sine, we get:
2(2sin^2(x/2) - 1) - 1 = 0
4sin^2(x/2) - 2 - 1 = 0
4sin^2(x/2) - 3 = 0
5. Simplify the equation further.
sin^2(x/2) = 3/4
6. Take the square root of both sides of the equation, considering both positive and negative values.
sin(x/2) = ±sqrt(3)/2
7. Now, solve for x/2 in the given interval [0, 2π).
To find the solutions, we need to determine the values of x/2 that satisfy sin(x/2) = ±sqrt(3)/2 within the given interval.
By using inverse trigonometric functions, we find:
x/2 = π/3, 2π/3, 4π/3, 5π/3
Now, multiply each value of x/2 by 2 to find the exact solutions for x.
x = π/3 * 2, 2π/3 * 2, 4π/3 * 2, 5π/3 * 2
The exact solutions of the equation sin(x/2) + cos(x) = 0 in the interval [0, 2π) are:
x = π/3, 2π/3, 4π/3, 5π/3.