We need to solve `2cos^2x-1=1`

Add 1 on both sides of the equation.

`2cos^2x=2`

Divide each term in the equation by 2.

`cos^2x=1`

Take the square root of both sides of the equation to eliminate the exponent on the left-hand side.

`cosx=+-1`

We have `x=cos^(-1)(1)` and `x=cos^(-1)(-1)`

Since we know that the graphs of trigonometric functions have repeating cycles, we get

`x= 0+-2Pi*n` and `x = PI+-2Pi*n` where `n ` is any integer.

The trigonometric equation `2cos^2x - 1 = 1` has to be solved.

`2cos^2x - 1 = 1`

=> `2cos^2x = 1 + 1`

=> `2cos^2x = 2`

=> `cos^2x = 1`

=> `cos x = +-1`

=> `x = cos^-1(1)` and `x = cos^-1(-1)`

=> `x = 0+-2*n*pi` and `x = pi+-n*2*pi`

**The solution of the equation `2cos^2x - 1 = 1` is `x = +-2*n*pi` and `x = pi+-n*2*pi` **

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