# complicated equation `arcsin(1-log_2(x^2+1))=arc cos(log_2(x^2+1))`

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### 1 Answer

You need take sine function both sides such that:

`sin(sin^(-1)(1-log_2(x^2+1))) = sin(cos^(-1)(log_2(x^2+1)))`

You need to remember that `sin(sin^(-1) x) = x` and `sin(cos^(-1) x) = sqrt(1 - x^2)` such that:

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

You need to raise to square both sides such that:

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

Expanding the binomial yields:

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

You need to move all terms to the left side such that:

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

You need to divide by 2 such that:

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

You should come up with the substitution `log_2(x^2+1) = y ` such that:

`y^2 - y = 0 =gt y(y - 1) = 0 =gt y_1 = 0 and y-1 = 0 =gt y_2 = 1`

You need to solve for x the equation `log_2(x^2+1) = y` such that:

`log_2(x^2+1) = y_1 =gt log_2(x^2+1) = 0 =gt x^2 + 1 = 2^0`

`x^2 + 1 = 1 =gt x^2 = 0 =gt x_1=x_2 = 0`

`log_2(x^2+1) = y_2 =gt log_2(x^2+1) = 1 =gt x^2+1 = 2 =gt x^2 - 1 = 0 =gt x^2 = 1 =gt x_(3,4) = +-1`

**Hence, evaluating the solutions to the given equation yields `x in {-1 ; 0 ; 1}.` **

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