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, ...