# Solve equation sin^-1(2^(x+2))-sin^-1(4squareroot3*2^x) = pie/2?

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You need to subtract `arcsin(4sqrt3*2^x))` such that:

`(arcsin(2^(x+2)) = pi/2 - arcsin(4sqrt3*2^x))`

You need to take sine function both sides such that:

`sin(arcsin(2^(x+2))) =sin(pi/2 - arcsin(4sqrt3*2^x)))`

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

`2^(x+2) = cos(arcsin(4sqrt3*2^x))`

`2^(x+2) = sqrt(1 - 48*2^(2x))`

You need to raise to square both sides such that:

`2^(2(x+2)) = 1 - 48*2^(2x)`

`16*2^(2x) + 48*2^(2x) = 1`

`64*2^(2x) = 1 =gt 2^(2x) = 1/64`

You need to write `64 = 2^6 =gt 1/64 = 1/(2^6) = 2^(-6)`

`2^(2x) =2^(-6) =gt 2x = -6 =gt x = -3`

**Hence, evaluating the solution the given equation yields `x = -3.` **