find all solutions to the equation in the interval [0,2pie]
cos2x=cosx
See:
http://www.jiskha.com/display.cgi?id=1303866873
cos 2x = cosx
2cos^2 x - 1 - cosx = 0
(2cosx + 1)(cosx -1) = 0
cosx = -1/2 or cosx = 1
if cosx = -1/2, x is in II or III
x = 2π/3 or 4π/3 ......I knew cos π/3 = +1/2
if cosx = 1, then x = 0 or x = 2π
To find all solutions to the equation cos(2x) = cos(x) in the interval [0, 2π], we can use trigonometric identities and solve for x.
Step 1: Apply the double-angle identity for cosine.
cos(2x) = 2cos²(x) - 1
Step 2: Substitute the expression back into the equation.
2cos²(x) - 1 = cos(x)
Step 3: Simplify the equation to solve for cos(x).
2cos²(x) - cos(x) - 1 = 0
Step 4: Factor the quadratic equation.
(2cos(x) + 1)(cos(x) - 1) = 0
Now, we set each factor equal to 0 and solve for cos(x):
Factor 1: 2cos(x) + 1 = 0
2cos(x) = -1
cos(x) = -1/2
Factor 2: cos(x) - 1 = 0
cos(x) = 1
Since cos(x) has a period of 2π, we need to find all the angles where cos(x) equals -1/2 and 1 within the given interval [0, 2π].
For cos(x) = -1/2:
One solution is x = π/3 in the interval [0, 2π].
Another solution is x = 5π/3 in the interval [0, 2π].
For cos(x) = 1:
One solution is x = 0 in the interval [0, 2π].
Another solution is x = 2π in the interval [0, 2π].
So, the solutions to the equation cos(2x) = cos(x) in the interval [0, 2π] are x = π/3, 5π/3, 0, and 2π.