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We need to solve `2cos^2x-1=1`
Add 1 on both sides of the equation.
Divide each term in the equation by 2.
Take the square root of both sides of the equation to eliminate the exponent on the left-hand side.
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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