# calculus

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Find all solutions to the equation in the interval [0,2pi)

Cosx-cos2x

• calculus -

I will assume your "equation" is
cosx - cos 2x = 0
cosx - (2cos^2 x - 1) = 0
2cos^2 x - cosx - 1 = 0
(2cosx + 1)(cosx - 1) = 0
cosx = -1/2 or cosx = 1

if cosx = -1/2 , then
x = 2π/3 or x = 4π/3 , (120° or 240°)
if cosx = 1 , then
x = 0 or 2π , (0° or 360°)

• calculus -

Thanks!

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