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We have to solve:

$ \sin(\dfrac{1}{2}x + \pi) $ = $\dfrac{1}{2}\sqrt{2}$

I get these answers:

$ x = -1\frac{1}{2}\pi + k2\pi\quad \lor\quad x =-\frac{1}{2}\pi + k2\pi$

Both these answers can be merged into this answer right?:

x = $\frac{1}{2}\pi + k\pi$

Also, are these 2 mergable (and have I merged them correctly or not?):

$ x = -\frac{1}{2}\pi+k6\pi\quad$ and $\quad x = 3\frac{1}{2}\pi+k6\pi$


$ x = 1\frac{1}{2}\pi+k3\pi$

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The second one is not correct. You can't merge them because they aren't half way around from each other: $3\frac 12\pi - (-\frac 12 \pi) \neq 3 \pi$ – Ross Millikan Sep 16 '12 at 13:19
up vote 1 down vote accepted

There is no value of $k$ for which $(1/2)\pi+2\pi k$ evaluates to $-(1/2)\pi+2\pi k$, so your merge doesn't work.

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But then x = $\frac{1}{2}\pi + k\pi$ would work then right? – JohnPhteven Sep 16 '12 at 12:09
Read my question again, I think now both answers should be correct.. – JohnPhteven Sep 16 '12 at 12:10
Can you please answer this? – JohnPhteven Sep 16 '12 at 12:59
Let's look at a simpler question, $\sin y=\sqrt2/2$. The solutions are $y=(\pi/4)+2k\pi$ and $y=(3\pi/4)+2k\pi$. I don't think there's any way to merge these into a single formula. Now set $(1/2)x+\pi=y$, then $x=2y-2\pi$, and combining this with the formulas for $y$ we get formulas for $x$. Why do you think you can merge them into a single formula? It's easy enough to check, anyway, by trying a few values of $k$ in each of the formulas to see whether they give the same set of values. Try it! – Gerry Myerson Sep 17 '12 at 7:17

$\sin(\frac{x}{2}+\pi)=\frac{\sqrt 2}{2}$


$\frac{x}{2}+\pi=t$ we have:

$\sin t=\frac{\sqrt 2}{2}$

The elementary trigonometrics equation is $\sin x=a$, $a\in R$. If $0<a<1$ solve the equation is $x=x_0+2k\pi$ and $x=\pi-x_0+2k\pi$ where $x_0=\arcsin|a|$.

$0<a=\frac{\sqrt 2}{2}<1$ $\Rightarrow$ $t_0=\arcsin\frac{\sqrt 2}{2}$ $\Rightarrow$ $t_0=\frac{\pi}{4}$.

Definitly: $t=\frac{\pi}{4}+2k\pi$, and $t=\frac{3\pi}{4}+2k\pi$.

From here we have:

$\frac{x}{2}+\pi=\frac{\pi}{4}+2k\pi$ $\Rightarrow$ $x=-\frac{3\pi}{2}+4k\pi$, and $\frac{x}{2}+\pi=\frac{3\pi}{4}+2k\pi$ $\Rightarrow$ $x=-\frac{\pi}{2}+4k\pi$, $k\in Z$

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I have edited, can you check if my new answers are correct? I think they are – JohnPhteven Sep 16 '12 at 12:59

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