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Use substitution to solve for $x$ in the following equation:

$$\frac{1}{2-\sin x}=\sin x$$

This is what I have done so far:

$$\sin^2x-2\sin x+1=0$$

$$\arcsin(1)=\frac{\pi}{2}=x$$

The correct answer is apparently:

$$x=\frac{\pi}{2}+2\pi n$$

I don't really understand why the $+ 2πn$ is there, could anyone advise me?

Also, I'm not really sure about the substitution aspect of the question. Should I have done something along the lines of letting $y=\sin x$ and then solving for $y$?

Thanks.

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  • $\begingroup$ $\sin(x)=1$ has more than one solution as sine is a periodic function $\endgroup$
    – Henry
    Jan 21 '15 at 20:04
  • $\begingroup$ It's true that $\sin x = 1$. But you didn't really use substitution. Writing $y = \sin x$ is what was meant by the instruction to use substitution. $\endgroup$
    – user208259
    Jan 21 '15 at 20:09
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The substitution seems unnecessary for this problem, but your intuition is right about using $y = \sin(x)$ should you have proceeded that way. Alternatively, you can complete the square $$\sin^2(x)-2\sin(x)+1 = 0 \\ \implies (\sin(x)-1)^2=0 \\ \implies \sin(x)-1 =0 \\ \implies \sin(x) = 1$$ Thinking back on the unit circle, you should know that $\sin(x) = 1$ when $x = \frac{\pi}{2}$. Remember, this corresponds to the polar coordinate $\left(1,\frac{\pi}{2}\right)$. However, you can also travel once around the unit circle (via adding $2\pi$) and you end up back at the same polar coordinate as $\left(1,\frac{\pi}{2}+2\pi\right) = \left(1,\frac{\pi}{2}\right)$. With that logic in mind, you can add or subtract multiples of $2\pi$ from $\frac{\pi}{2}$ and you will still be at the same coordinate. Hence for all $n \in \Bbb{Z}$, $$\left(1,\frac{\pi}{2}+2n\pi\right) = \left(1,\frac{\pi}{2}\right)$$ and thus $\sin(x) = 1$ when $x = \frac{\pi}{2},\space \frac{\pi}{2} \pm 2\pi, \space \frac{\pi}{2} \pm 4\pi, \ldots$

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You have done everything right except for the last step when solving the equation $\sin(x)=1$. There are actually infinitely many solutions and they are $\frac{\pi}{2}+2\pi n$ where $n$ is any integer.

The problem occurred when you applied the $\arcsin$ function to solve the equation. The $\arcsin$ function is defined to be the inverse of $\sin$ but only on the interval $[-\frac{\pi}{2},\frac{\pi}{2}]$ since otherwise it would not be a function because inputs would have multiple outputs. However, in your case you want to find all the possible $x$ such that $\sin x=1$.

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You found that $\sin(x) = 1$, that is great. Also you know that $\sin(\pi/2) = 1$ so $\pi/2$ is a solution, but is not the only solution.

You can use the trigonometric circle to visualize, $\pi/2$ is on the top, but if you make one more turn (clockwise or not) you get another solution, which will be $\pi/2 + 2\pi$ or $\pi/2 - 2\pi$. You can make more turns, as much as you want. This way you get infinite solutions, all in the form $\pi/2 + 2\pi n$, with $n\in\mathbb{Z}$. They are all solutions, because it is only on the top of the trigonometric circle that you have $\sin(x) = 1$.

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what you have is right. if you insist on substitution, then we can substitute $y = \sin x$ in $\dfrac{1}{2 - \sin x} = \sin x$ and turn into $\dfrac{1}{2-y} = y.$ now we need to recognize this as a quadratic equation in $y$ by cross multiplying and rearranging to get $$y^2 - 2y + 1 = 0$$ which has a repeating solution $y = 1.$

going back to the old variable $x,$ we need to solve $\sin x = 1$ the only solution for $x$ in one period $[0, 2\pi]$ is $x = \pi/2$ if you are looking for all solutions, you can multiples of the period, $2\pi,$ to the solution and get $$x = \pi/2 + 2k\pi, \text{ where $k$ is an integer} $$

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