# Solve the equation `sin^-1(1-log_2 x)=cos^-2(log_2 x)`

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The equation `sin^-1(1 - log_2x) = cos^-1(log_2x)` has to be solved. Use the relation `cos^-1x = sin^-1 sqrt(1 - x^2)`

Let `log_2x = y`

`sin^-1(1 - log_2x) = cos^-1(log_2x)`

=> `sin^-1(1 - y) = sin^-1(1 - y^2)`

=> `1 - y = 1 - y^2`

=> `y^2 - y = 0`

=> `y(y - 1) = 0`

=> y = 0 and y - 1 = 0

y = 0 and y = 1

`log_2 x = 0`

=> x = 1

`log_2 x = 1`

=> x = 2

**The solution of the equation is x = 1 and x = 2**

You should remember that `sin(arcsin x) = x` and `sin(arccos x) = sqrt(1 - x^2)` , hence, reasoning by analogy yields:

`sin(arcsin(1-log_2 x)) = sin(arccos(log_2 x))`

`1 - log_2 x = sqrt(1 - log_2^2 x)`

You need to raise to square to remove the square root such that:

`(1 - log_2 x)^2 = (1 - log_2^2 x)`

You should move all terms to one side such that:

`(1 - log_2_2 x)^2- (1 - log_2^2 x) = 0`

You need to factor out `(1 - log_2^2 x)` such that:

`(1 - log_2^2 x)(1 - log_2^2 x - 1) = 0`

Reducing like terms yields:

`(1 - log_2^2 x)(log_2^2 x) = 0`

`log_2^2 x = 0 => x = 2^0 => x = 1`

`1 - log_2^2 x = 0 => - log_2^2 x = -1 => log_2^2 x = 1 => 1 - log_2 x = +-1 =>` `x = 2` or `x = -2`

You need to keep the positive values since if x is negative, the logarithm is invalid.

**Hence, evaluating the solutions to the given equation yields `x = 1` and `x = 2` .**