# How to solve these types of trig equations

Lets use an example:

$$\sin^2 \left(\dfrac{\pi}4x\right) = 1$$

I am at this point:

$$\frac{\pi}4 x=\frac{\pi}2 + k\cdot2\pi \quad\text{or}\quad \frac{\pi}4 x=-\frac{\pi}2 + k\cdot2\pi$$

But then you have to merge the formulae into $\frac{\pi}4 x = \frac{\pi}2 + k\cdot\pi$

This is not a hard example, but I have a LOT of trouble knowing when they are and when they aren't 'mergeable' , and how to easily figure out how to merge 2,3 or even 4 of these formulae into 1. How can I make this less troublesome? I am certain there is an easier way than just plain figuring it out in your head.. (like for example, I have no idea how to know quickly if for example $x =\frac{\pi}2 + k\cdot2\pi$ and $x=k\cdot2\pi$ are mergeable (my textbook indicates they aren't))

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Does $1/4 \pi x$ mean $\dfrac{\pi x}{4}$ or $\dfrac{1}{4\pi x}$? – Henry Sep 15 '12 at 11:15
The first one. How do you do that? – JohnPhteven Sep 15 '12 at 11:24
$\dfrac{\pi x}{4}$ – Henry Sep 15 '12 at 11:33
$\dfrac{\pi x}{4}$ Wow that is pretty complicated :P – JohnPhteven Sep 15 '12 at 11:40
It is not as bad as <math xmlns="w3.org/1998/Math/MathML">; <mstyle displaystyle="true"> <mfrac> <mrow> <mi>&#x03C0;<!-- π --></mi> <mi>x</mi> </mrow> <mn>4</mn> </mfrac> </mstyle> [/itex] – Henry Sep 15 '12 at 13:39

## 1 Answer

To avoid confusion, try introducing $t\equiv \frac{\pi x}{ 4}$.

You obtain $\sin^2t=1 \longrightarrow \sin t=\pm 1$, which are two easy to solve equations.

Once you have done this, substitute every solution in the equality that defines $t$ and solve for $x$.

About merging the solutions. First of all, you don't need to do that: saying that the full solution is the union of the partial solutions is perfectly fine. But I agree with you that sometimes it isn't so elegant.
Then you have to train your eye, so that you can notice patterns in the partial solutions, such that you can present them in a more compact way. Sometimes it's just unuseful to try this exercise (e.g. if you solve $\sin x = \frac12$, don't worry about merging $x=\frac{\pi}6+2k\pi$ and $x=\frac{5\pi}6+2k\pi$)!

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I understand, but I don't know how to merge those formulae. I some cases you have 2 very different formula which still can be merged into one. My question is; how can you figure out of 2 formulae can be merged into one? – JohnPhteven Sep 15 '12 at 11:19
Excuse me, I didn't understand your solution and thought it had a more serious flaw. I edited to address your question and I'll also edit your first post, so that it isn't ambiguous. :) – Andrea Orta Sep 15 '12 at 11:27
@ Andrea Yes I too find it a bit redundant, however, I think we HAVE to merge the answers, our teacher is very strict in this. Also, if you merge my example like I did in the question, your final answer is $x = 2 + k . 4$ However, if you don't merge it I get these 2 answers: $x = 2 + k . 8$ and $x = -2 + k . 8$ So I get 2 different answers.. – JohnPhteven Sep 15 '12 at 11:29
You don't get two different answers: you get two partial answers, both correct (i.e. they both satisfy the equation), but you have to take their union! Anyway, to merge $x=\pm2 + 8k$ in $x=2+4k$ you have to see it: the only important thing to understand is the role of that $k$. $k$ is free to be any integer number (also negative, also $0$): once you have merged two partial solutions, try letting $k$ be different numbers and check if you obtain all the values of your partial solutions (and only those values!). Otherwise, you made a mistake in the merging process. – Andrea Orta Sep 15 '12 at 11:44
Hmm, I see it now, those 2 answers can be merged into $x = 2 + 4k$ But now I have another question. What does $x = 2 + 4k$ even mean? It has no pi in it, does it just mean like in radians, not in pi radians? That is my last question, a lot of thanks for your help! – JohnPhteven Sep 15 '12 at 11:48