How to solve $\cos x+i\sin x=\sin x+i\cos x$ where $x\in\Bbb{R}$ using basic properties

Question

Solve $$\cos x+i\sin x=\sin x+i\cos x$$ where $x\in\Bbb{R}$ and graph the complex region.

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I know that $$e^{ix}=\cos x+i\sin x$$ and returning to the original equation: $$e^{ix}=\sin x+i\cos x,$$ but I am not able to find why $$\sin x+i\cos x=ie^{-ix}$$ (according to WolframAlpha).

We know that $\cos x=\cos(-x)$ and $\sin x=-\sin(-x)$, hence if $e^{ix}=\cos x+i\sin x$ then $e^{ix}=\cos(-x)-i\sin(-x)$ then $ie^{ix}=\sin(-x)+i\cos(-x)$, but then?

The answer should be $$x=\frac{\pi}{4}+k\pi,\;k\in\Bbb{Z}$$ but I am not able to reach it. Any help?

Answer 1

If

$\cos x + i\sin x = \sin x + i \cos x, \tag 1$

then equating real and imaginary parts we find

$\cos x = \sin x; \tag 2$

since $\cos x$ and $\sin x$ cannot be simultaneously zero, we infer

$\cos x \ne 0 \ne \sin x; \tag 2$

we may thus write

$\tan x = \dfrac{\sin x}{\cos x} = 1; \tag 3$

the values of $x$ for which this binds are

$x = \dfrac{\pi}{4} + k\pi, \; k \in \Bbb Z; \tag 4$

these are then the solutions to (1).

I leave graphing these results to my readership.

Further, with

$e^{ix} = \cos x + i\sin x, \tag 5$

we have

$e^{-ix} = e^{i(-x)} = \cos (-x) + i\sin (-)x = \cos x - i\sin x; \tag 6$

thus,

$i e^{-ix} = i\cos x - i^2\sin x = i\cos x + \sin x. \tag 7$

Answer 2

Write it as $\cos x-\sin x+i(\sin x-\cos x) =0 $.

Since $x \in \mathbb{R}$, the real and imaginary parts must each be zero, so $\cos(x) = \sin(x) = \cos(\pi/2-x) $.

Since $\cos$ has period $2\pi$, $x-(\pi/2-x) = 2k\pi $ so $2x = 2k\pi+\pi/2 $ or $x = k\pi+\pi/4 $.

Answer 3

This is one you can do by inspection. One thing you might wonder is when sin(x) = cos(x), because this would be what would solve this! No need for anything more complex than that.