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Take the cosine of both sides to get
cos(2arccos(x))=2(cos(arccos(x)))^2-1 and cos(arccos(2x-1))=2x-1 and
cos(arccos(x))=x so we get
2x^2 - 1 = 2x - 1
2x^2 - 2x = 0
Factoring we get
2x(x-1) = 0
So x = 0 or x = 1 are our solutions. You should substitute into the original equation to check the answers.
2arccos(0) = arccos(-1) which arccos(0) = pi/2 and arccos(-1) = pi so this checks.
2arccos(1) = arccos(1) since arccos(1) = 0 this checks.
So the answers are x = 0 and x = 1.
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