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

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]}

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#