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The assignment is that I'm suppose to correct a flawed solution to the equation $\sin^2 x = 1$. The flawed solution is:

$\sin^2 x = 1$

$\sin x = 1$

$x = 90^\circ + 2n\pi$

I thought I was simply to correct the fact that they forgot the negative root. So my solution was, skipping the solving steps here obviously:

$x = 2n\pi - 90^\circ$

However my teacher says that I am repeating an error that was also present in the flawed solution. Apparently it had something to do with the last line in the original solution. I have also stated that $n$ is an arbitrary integer.

Any help is greatly appreciated!

EDIT:

I think the error the teacher is referring to is that I'm expressing the solution a bit weird by mixing radians and degrees, like you guys mentioned. I've also defined both roots more clearly like @amWhy did below.

Thanks guys!

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    $\begingroup$ Probably not good to mix radians and degrees. $\endgroup$ – Thomas Andrews Mar 29 '13 at 18:54
  • $\begingroup$ You are absolutely right, I will stick to radians for this solution. Thanks! $\endgroup$ – Johan Mar 29 '13 at 19:15
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Hint: Your solution omitted what the original solution conveyed, and the original solution omitted what your solution conveyed:

You need to find solutions for both roots of $\sin^2 x = 1$:

$$\sin^2 x = 1 \iff \sin x = 1 \;\;\text {or} \;\;\sin x = -1$$

Any $x$ satisfying either of the equations is a solution to the given equuation.


Also note that both the original solution and your solution: $\pm 90^\circ + 2n\pi\;$ mixes degrees with radians; by convention, we express the solutions with the consistent use of units.

$$x = 2n\pi \pm \frac{\pi}{2}\quad n \in \mathbb Z \quad\quad \text{or, equivalently} \quad \quad x= {k\pi} + \pi/2 \quad k \in \mathbb Z $$

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  • $\begingroup$ I might be completely off here, I have had a glass or two... But how can $x = 2n\pi +- \pi/2 = ((2n +1)\pi)/2$ Since if I use an arbitrary number, say $n = 1$. Then $ 2*1*\pi + \pi/2 =(approx) 7.85$.... However $((2*1+1)\pi)/2 =(approx) 4.7...$ I'm lost $\endgroup$ – Johan Mar 29 '13 at 19:10
  • $\begingroup$ Okay, we need both solutions: $x = 2n\pi - \pi/2$ (solves $\sin x = -1$) AND $x = 2n\pi + \pi/2$ (solves $\sin x = 1$). These solutions can both be covered by the solution $\dfrac{(2n+1)\pi}{2}$. Your corrected solution solves only for $\sin x = -1$, and the original solution solves only for $\sin x = 1$. You need them both. $\endgroup$ – Namaste Mar 29 '13 at 19:46
  • $\begingroup$ Does this make sense now? If you're uncomfortable with the "general solution", just solve for each root, and include both solutions, in radians. ;-) $\endgroup$ – Namaste Mar 29 '13 at 19:50
  • $\begingroup$ Allright cool, however another valid solution to ($\sin x = -1$ and $\sin = 1$ would be the one you posted above $x = 2n\pi \pm \frac{\pi}{2}$ right? Thanks for your help man, I really appreciate it! $\endgroup$ – Johan Mar 29 '13 at 19:52
  • $\begingroup$ Yes, absolutely. With the $\pm$, you cover solutions to both roots. $\endgroup$ – Namaste Mar 29 '13 at 20:25
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$\cos^2x=1-\sin^2x=0\implies \cos x=0\implies x=\frac{(2n+1)\pi}2$ where $n$ is any integer


Alternatively,

$\sin^2x=1\implies \cos2x=1-2\sin^2x=1-2\cdot 1=-1=\cos\pi$

So,$2x=2n\pi+\pi=(2n+1)\pi$ where $n$ is any integer

So, $x=\frac{(2n+1)\pi}2$

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One way to avoid the $\pm 1$ issue is to notice that $\cos^2 x + \sin^2 x = 1$. Then, solutions exist when $\cos^2 x = 0$, which is exactly when $\cos x = 0$.

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  • $\begingroup$ A downvote? Seriously? $\endgroup$ – Emily Mar 30 '13 at 23:26

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