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I have to inverse these trigonometric functions: $f(x)=\sin(2x), \space x\in[-\frac{\pi}{4},\frac{\pi}{4}]\\, f(x)=\sin(3x)-3, \space x\in[-\frac{\pi}{6},\frac{\pi}{6}]\\, f(x)=2\sin(3x+\frac{\pi}{4}), \space x\in[-\frac{3\pi}{4},\frac{\pi}{4}]\\, f(x)=2 - \sin(4x), \space x\in[-\frac{\pi}{8},\frac{\pi}{8}]$,

The problem is that I'm unsure what to do with the domain. Should I fiddle with it so that It matches $\arcsin$ domain of $[-\frac{\pi}{2}, \frac{\pi}{2}]$? And if so, how do I do that? I came up with the solution for the first one but I'm unsure if it's correct: $f^{-1}(x)=\frac{\arcsin(x)}{2}$

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hint for 1,2

Use the fact that

$$\sin(X)=Y \;and\; X\in[-\frac{\pi}{2},\frac{\pi}{2}]\iff X=\arcsin(Y)$$

So, $$f(x)=\sin(2x)\;and\; 2x\in[-\frac{\pi}{2},\frac{\pi}{2}]\iff 2x=\arcsin(f(x))$$

$$\iff x=\frac 12\arcsin(f(x))$$

$$f^{-1}(x)=\frac 12\arcsin(x).$$

For the second, you should find $$f^{-1}(x)=\frac 13(x+3)$$

For the fourth, you will get $$f^{-1}(x)=\frac 14(2-x)$$ The third is not a bijection since $$f(-\frac{3\pi}{4})=f(\frac{\pi}{4})=0$$

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