Question

Is there a value of `x` such that `sin x = 3/2` ?

Answer 1

No

Explanation:

Within the realm of real numbers (your standard 1,2,3.1,pi,0 etc), `sinx` cannot be `3/2` for any x.

This is because the sin function will only output numbers ranging from `-1` to `1`.

Visually, we can show this by graphing `y=sin(x)` , which clearly never reaches a height of `3/2`

graph{sin(x) [-10, 10, -5, 5]}

Answer 2

Yes, but only for complex values of `x`

`x = -i ln (1/2(3+-sqrt(5))i) + 2npi" "n in ZZ`

Explanation:

As a real valued function of real numbers, `sin x` maps `(-oo, oo)` onto `[-1, 1]`, so does not take the value `3/2`.

So there is no real number `x` such that `sin x = 3/2`

However, consider the following:

`e^(itheta) = cos theta + i sin theta`

So:

`e^(itheta) - e^(-itheta) = (cos theta + i sin theta) - (cos(-theta) + i sin(-theta))`

`color(white)(e^(itheta) - e^(-itheta)) = (cos theta + i sin theta) - (cos(theta) - i sin(theta))`

`color(white)(e^(itheta) - e^(-itheta)) = 2i sin theta`

So we find:

`sin x = (e^(ix)-e^(-ix))/(2i)`

We can use this definition of `sin x` for complex values of `x`.

Then we want to solve:

`3/2 = sin x = (e^(ix)-e^(-ix))/(2i)`

Multiply both ends by `2` to get:

`3 = e^(ix)/i - 1/(ie^(ix)) = e^(ix)/i + i/e^(ix) = t+1/t`

where `t= e^(ix)/i`

Multiply both ends by `4t` to get:

`12t = 4t^2+4`

Subtract `12t` from both sides to get:

`0 = 4t^2-12t+4`

`color(white)(0) = (2t)^2-2(2t)(3)+3^2-5`

`color(white)(0) = (2t-3)^2-(sqrt(5))^2`

`color(white)(0) = ((2t-3)-sqrt(5))((2t-3)+sqrt(5))`

`color(white)(0) = (2t-3-sqrt(5))(2t-3+sqrt(5))`

Hence:

`e^(ix)/i = t = 1/2(3+-sqrt(5))`

So:

`e^(ix) = 1/2(3+-sqrt(5))i`

So:

`ix = ln (1/2(3+-sqrt(5))i) + 2npii" "n in ZZ`

So:

`x = -i ln (1/2(3+-sqrt(5))i) + 2npi" "n in ZZ`