Is there a value of #x# such that #sin x = 3/2# ?
2 Answers
No
Explanation:
Within the realm of real numbers (your standard 1,2,3.1,pi,0 etc),
This is because the sin function will only output numbers ranging from
Visually, we can show this by graphing
graph{sin(x) [-10, 10, -5, 5]}
Yes, but only for complex values of
Explanation:
As a real valued function of real numbers,
So there is no real number
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
Then we want to solve:
#3/2 = sin x = (e^(ix)-e^(-ix))/(2i)#
Multiply both ends by
#3 = e^(ix)/i - 1/(ie^(ix)) = e^(ix)/i + i/e^(ix) = t+1/t#
where
Multiply both ends by
#12t = 4t^2+4#
Subtract
#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#