# Find the inverse of $f(x)=\sin(3x-1)$ over the domain $\left[\frac{2-\pi}{6},\frac{2+\pi}{6}\right]$

So I'm preparing for my further studies (last year of high school, preparing so I can try and join the academy that I want), and just solving problems. Got stuck on this one:

Find the inverse of $$f(x)=\sin(3x-1)$$ in the domain of $$x \in \left[\dfrac{2-\pi}{6}, \dfrac{2+\pi}{6}\right]$$.

What I've tried so far is:

$$f(x) = y\\y=\sin(3x-1)\\y = \sin(3x)\cos(1)-\cos(3x)\sin(1)$$

At this point I have no idea what to do, I thought of trying to split $$3x=2x+x$$ and continuing but it would be too messy.

You are probably over thinking about the simple thing.

Recall the definition of Inverse of a function.

In mathematics, an inverse function (or anti-function) is a function that "reverses" another function: if the function f applied to an input x gives a result of y, then applying its inverse function g to y gives the result x, and vice versa, i.e., f(x) = y if and only if g(y) = x.

To write an inverse of a function, you have to represent the whole function of $$x$$ in $$f(x)$$ or $$y$$. What I mean from this is:

$$f(x) = sin(3x - 1)$$

You can call $$f(x)$$ as $$y$$.

$$y = sin(3x - 1)$$

Now write the whole function of x in terms of y.

$$\sin^{-1}(y) = 3x - 1$$

$$(\sin^{-1}(y) + 1)/3 = x$$

And this is your inverse function.

Now if you want to write it in terms of π or more precisely, you can write 1 as $$(\sin^{-1}(π/2)$$ and apply $$sin(A) + sin(B)$$.

• Yeah I knew how to get the inverse of a function, the problem was I didn't know what to do with the y=sin(3x-1), got it as arcsin(y) = 3x-1, and just solved that for x Oct 7, 2018 at 10:29
• That's great! If this answer is correct according to your question, Kindly mark it as correct and upvote it. I'd appreciate that. Thank you. Oct 7, 2018 at 10:34

Hint: If $$y=\sin(3x-1)$$ then $$\arcsin(y)=3x-1$$

HINT:

• Rearrange the equation so you have $$x$$ in terms of $$f(x)$$

• Write down a domain for this inverse function, by considering the possible values of $$\arcsin$$

The equation $$\sin(3x-1)=y$$ has the solutions $$3x-1=\arcsin y+2k\pi \qquad\text{or}\qquad 3x-1=\pi-\arcsin y+2k\pi$$ Since you know that $$\frac{2-\pi}{6}\le x\le\frac{2+\pi}{6}$$, you also have $$3\frac{2-\pi}{6}-1\le 3x-1\le 3\frac{2+\pi}{6}-1$$ that is, $$-\frac{\pi}{2}\le 3x-1\le\frac{\pi}{2}$$ so we have to take the solution from the first family, with $$k=0$$, because by definition $$-\pi/2\le\arcsin y\le\pi/2$$.

Thus $$3x-1=\arcsin y$$ and therefore $$x=\frac{1}{3}(1+\arcsin y)$$ The inverse is the function, defined over $$[-1,1]$$, $$g(y)=\frac{1}{3}(1+\arcsin y)$$

Different domains for $$\sin(3x-1)$$ would lead to different determinations of the family of solutions or of $$k$$. For instance, the inverse of $$f(x)=\sin(3x-1)$$ over $$\bigl[\frac{\pi+2}{6},\frac{3\pi+2}{6}\bigr]$$ would be $$g(y)=\frac{1}{3}(1+\pi-\arcsin y)$$