# What is a value of x that satisfies the equation 2cosx - cosx-1=0

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You need to collect like terms such that:

`(2cos x - cos x ) - 1 = 0`

You need to keep the cosines to the left side and you should move the constant term to the right side such that:

`cos x = 1`

You need to remember that the cosine function is positive in quadrants I and IV, hence:

`x = cos^(-1) 1 + 2n*pi`

`x = 2n*pi`

You need to remember that the period of cosine function is 2pi, hence the value of 1 will repeat every `2pi` radians, or every `360^o` .

**Hence, the number of solutions to the equation is infinite and the solutions have the following form `x=2npi, n in Z` set.**

2cos (x) - cos (x) -1 = 0

cos (x) = 1

x = 0, 2pi, ...