Find all solutions in the interval[0,2pi),cos x/2-sinx=0
To find all solutions in the interval [0, 2π) for the equation cos(x/2) - sin(x) = 0, you can follow these steps:
Step 1: Use the trigonometric identity cos(2θ) = 1 - 2sin^2(θ) to rewrite the equation in terms of a single trigonometric function.
cos(x/2) - sin(x) = 0
cos(x/2) = sin(x)
Step 2: Further simplify the equation by using the double angle identity sin(2θ) = 2sin(θ)cos(θ).
cos(x/2) = 2sin(x/2)cos(x/2)
Step 3: Apply the trigonometric identity cos(θ) = sqrt(1 - sin^2(θ)).
cos(x/2) = 2sin(x/2)sqrt(1 - sin^2(x/2))
Step 4: Square both sides of the equation to eliminate the square root.
cos^2(x/2) = 4sin^2(x/2)(1 - sin^2(x/2))
Step 5: Apply the trigonometric identity cos^2(θ) = 1 - sin^2(θ).
1 - sin^2(x/2) = 4sin^2(x/2)(1 - sin^2(x/2))
Step 6: Expand and simplify the equation.
1 - sin^2(x/2) = 4sin^2(x/2) - 4sin^4(x/2)
1 = 5sin^2(x/2) - 4sin^4(x/2)
Step 7: Rearrange the equation to have it in the standard quadratic form.
4sin^4(x/2) - 5sin^2(x/2) + 1 = 0
Step 8: Let u = sin^2(x/2) and replace sin^2(x/2) with u.
4u^2 - 5u + 1 = 0
Step 9: Solve the quadratic equation for u by factoring or using the quadratic formula.
(4u - 1)(u - 1) = 0
From this, we get two possible values for u: u = 1/4 and u = 1.
Step 10: Solve for x by substituting the values of u back into sin^2(x/2) = u and then solving for x.
sin^2(x/2) = 1/4:
(x/2) = arcsin(1/2) or (x/2) = π - arcsin(1/2)
For the first case, x would be: x = 2arcsin(1/2) or x = π - 2arcsin(1/2)
sin^2(x/2) = 1:
(x/2) = π/2
From this, we get another solution: x = π
Therefore, the solutions in the interval [0, 2π) for cos(x/2) - sin(x) = 0 are x = 2arcsin(1/2), π - 2arcsin(1/2), and π.