Can you solve the equation sin x = 2?

Preliminary

Is there any solution for the equation sin x = 2? The answer is yes if you accept complex numbers. However, the reader should know a bit of complex numbers and trigonometry.

Let us start with Euler’s Identity:

e^ix = cos x + i sin x (1)

Replace x by –x, and note that sin(-x) = - sin x, we get :

e^-ix = cos x - i sin x (2)

(1) – (2), e^ix – e^-ix = 2i sin x

We finally get:

Weird Equation

We start with sin z = a , where a > 1

Here we use the complex variable z, for we know that the equation has no real solution.

Putting z = x + iy, the equation becomes:

sin (x + iy) = a , a > 1.

However,

Setting this equal to the real number a, and compare real and imaginary part, we have:

From (5),

cos x = 0 or e^y  - e^-y = 0

x = p/2 or y = 0

However, if y = 0, then sin x = a and this works well only if a £ 1.

\ x = p/2, and we have sin x = 1.

From (4), we have e^-y + e^y = 2a

e^2y –2a e^y + 1 = 0

This is a quadratic equation, solving it will get:

Readers please check that the ± sign can move out of the ln sign!

Finally, we get the formula:

More points of interest

(1) If a < -1, the formula is:

(2) Here are some equations, you may verify their solutions:

Equation	Solutions
sin x = 1	1.5708
sin x = 2	1.5708 ± 1.31696 i
sin x = 3	1.5708 ± 1.76275 i
sin x = -2	- 1.5708 ± 1.31696 i
sin x = i	± 0.881374 i