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

2 Answers
Sep 27, 2017

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]}

Sep 27, 2017

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