# Math Help - Trigonometric equations and their solutions

1. ## Trigonometric equations and their solutions

π=pi - format looks like an n =/

sinx=1/2

x=π/6 and x=π-π/6=5π/6

x=π/6+2nπor x=5π/6+2nπ

now the next problem does:

sinx=-1/2
x=π+π/6=7π/6 and x=2π-π/6=11π/6

x=7π/6+2nπ and x=11π/6+2nπ

now why does the first problem uses π as oppose to 2π I don't know what i'm suppose to subtract/add to in order to have the solution. When i do the homework it's always the other and i don't understand why. Am i suppose to use the angle, π, or 2π? The first problem uses it's angle and π - it's angle while the second problem uses π + its angle and 2π - it's angle to get a solution.

2. Originally Posted by Frenchie
π=pi - format looks like an n =/

sinx=1/2

x=π/6 and x=π-π/6=5π/6

x=π/6+2nπor x=5π/6+2nπ

now the next problem does:

sinx=-1/2
x=π+π/6=7π/6 and x=2π-π/6=11π/6

x=7π/6+2nπ and x=11π/6+2nπ

now why does the first problem uses π as oppose to 2π I don't know what i'm suppose to subtract/add to in order to have the solution. When i do the homework it's always the other and i don't understand why. Am i suppose to use the angle, π, or 2π? The first problem uses it's angle and π - it's angle while the second problem uses π + its angle and 2π - it's angle to get a solution.
Hey Frenchie,

Think of the graph of a trigonomic equation, its a wave that passes the x-axis over and over, and infinite amount of times. Each hit on the x-axis is a solution to the equation, so there is infinitely many solutions. So you may have come up with one solution, but instead the problem tells you that its a different solution. Each solution is only an additon of the period away from the first solution. The "period" of a trigonomic function is the distance between each "hit" on the x-axis (or the distance between the tops or bottoms of each "wave"). Let me show you with a general equation:

$f(x) = a \cdot sin(bx+c) + d$

Now, the period of the function is given by:

$P = \frac{2 \pi}{b}$

So, say we have the equation:

$f(x) = sin(4x)$

In the above equation its easy to see, and to check, that:

$a = 1$

$b = 4$

$c = 0$

and

$d = 0$

So, looking at this equation, we can find the period to be:

$P = \frac{2 \pi}{b} = \frac{2 \pi}{4} = \frac{2 \pi}{(2)(2)} = \frac{2}{2} \cdot \frac{\pi}{2} = (1)(\frac{\pi}{2}) = \frac{\pi}{2}$

So, looking at the equation again:

$f(x) = sin(4x)$

We can find the first solution. I assume you know that:

$sin(0) = 0$

So, we can see that if we let $x=0$ we get:

$sin(4 \cdot 0) = sin(0) = 0$

Which means that that is a "hit" on the x-axis, and that is the "first" solution. But its considered a trivial solution. But, now that you have this solution, you can find any other non-trivial solutions. If we let the first solution (or any solution for that matter) be represented by $r$, then any other solution (for sine and cosine functions) can be found with the following equation (where $x$ denotes another solution):

$x = r + nP$

$n$ must always be a whole number, in other words:

$n = ... -3, -2, -1, 0, 1, 2, 3 ...$

So, our first solution was $r = 0$

Now, we can figure out the first non-trivial solution with the equation:

$x = r + nP$

Lets let $n=1$, we know the period is $P = \frac{\pi}{2}$ so:

$x = r +nP = 0 + (n)(\frac{\pi}{2}) = 0 + (1)(\frac{\pi}{2}) = \frac{\pi}{2}$

Does that make sense?

Now, see if you can't solve the problems with that information. If not, post back here specificaly stating which problems you'd like some help with, and I'd be glad to assist. I hope this helps

3. looking at this let's try sinx=-1/2

-1/2=-π/6

so solution one would be -π/6 and solution two would be -π/6+2π=11π/6

The book shows the solution such as x=π+π/6=7π/6 and x=2π-π/6=11π/6

I don't know where solution one came from to begin with...

also for sinx=1/2 wouldn't the period still be 2π? cause it shows it as π.... i am so confused.

4. Originally Posted by Frenchie
looking at this let's try sinx=-1/2

-1/2=-π/6

so solution one would be -π/6 and solution two would be -π/6+2π=11π/6

The book shows the solution such as x=π+π/6=7π/6 and x=2π-π/6=11π/6

I don't know where solution one came from to begin with...
Generally, the way that it is taught to find solutions to trigonomic equations is to use "The Unit Circle" and "Special Right Triangles". Its much easier to explain these topics with a picture; heres the unit circle (you might have seen this before):

Each ordered pair in the above picture gives you a right triangle. The line outward to the point is the hypotenues, and the first number of the ordered pair is the x location of the point, so that would be the horizontal distance of the bottom side of the triangle, a.k.a. the length of that side; while the second number in the ordered pair is the y location of that point, so that would be vertical distance other side fo the triangle, a.k.a. the length of that side. Now, the number under each of the lines is the angle that this triangle makes with the x-axis. Something else you might have already seen, which is also used to solve trigonomic equations, is the "special" right triangles, pictured below:

Now, lets use your problem as an example:

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

You said you didn't know where the first solution came from, let me explain. So we know that the sine of an angle gives you the ratio of the opposite $(O)$ side (opposite to the angle $x$ that is) to the hypoteneous $(H)$; ) in a right triangle, and in the unit circle the angle is made with the x-axis always. So $sin(x) = \frac{O}{H}$. In your problem we can see that since:

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

then we know that the triangle we are looking for in the unit circle has:

$Opposite \; Side = O = -1$

$Hypotenues = H = 2$

or

$Opposite \; Side = O = 1$

$Hypotenues = H = -2$

We want to find the angle that results in these sides. If you memorized the special right triangles, you'd know that a ratio of 1 to 2 can be found in the 30, 60, 90 triangle, as seen in the picture. The three sides in this triangle are:

$1 = side \; opposite \; 30 \; degree \; angle$

$2 = side \; opposite \; 90 \; degree \; angle$

$\sqrt{3} = side \; opposite \; 60 \; degree \; angle$

Now, if we make a triangle that has a 30 degree angle with the x-axis (or a angle of $(\pi / 6)$ to be correct) and has a hypotenues of 2, then we know that the opposite side is $1$. So $sin(\pi/6) = 1/2$. But heres the thing, we need a negitive 1/2 to be equal to the sin of our angle. So, to make this happen, consider if we used one of the points in the quadrant where y is negitive and x is negitive. If we make the same angle with the x-axis we get the same triangle, except with a -1 instead of just 1 (we are only using the y distance, since this triangle makes its angle with the x-axis, and with sine we only use the opposite side and the hypoteneous, we only use the y distance of the point and the hypotenoues, and the hypotenues is always positive). But, and here is where it might get confusing, to get to this triangle, we "spun the dial" around the unit circle for some distance. If we started where x and y are both positive, and moved around counter clockwise, you can see that to create a triangle with the x axis that has a negitive vertical side (and thus looks upside down) we had to spin the dial past the angle measure of $\pi$. But, how far past did we go? Well that simple, $\pi/6$, since thats the angle made with the x-axis by the triangle! So, to find our angle, we need to add $\pi/6$ to the measurement of the angle we used to get back to the x-axis, being on the negitive x side. Like I said though, this angle is simply $\pi$, so $\pi + \pi/6$ is the answer!

Generally, questions like these are looking for answers in the range of $0 \leq x \leq 2\pi$, so although there are many more solutions (you simply "turn the dial" all the way around to get a new solution angle with your problem), they will generally be looking for the one less then $2\pi$ but greater then $0$. Now, does the process above make sense to you? Do you see how we arrived at:

$sin(\pi + \frac{\pi}{6}) = sin(\frac{7\pi}{6}) = \frac{-1}{2}$

If you have any more questions, or just plain didn't understand anything I explained, don't hesitate to post back here and ask more questions, or point out a place where you didn't understand what I meant, and I'd be happy to explain. I hope this helps

5. what i don't follow is that if we're looking for an angle that will be 30 by turning the "dial" aren't they all 30 degrees?(pi/6, 5pi/6, 7pi/6, 11pi/6). It's just a matter of using pi or 2pi to get that.... so i still dont understand how i figure out which is correct in order to get my solutions...

Sorry you probably made it clear with all the information provided above but i still don't understand =(

6. oh wait maybe i got it... so for sinx=1/2 it would be 5pi/6 because it's also positive 1/2 on the y axis and again pi/6 because it's positive 1/2. Since they are less than 270 degrees we would only use pi on the second one to get 30 degrees and nothing on the first solution because it's already a 30 degree angle?

7. Originally Posted by Frenchie
oh wait maybe i got it... so for sinx=1/2 it would be 5pi/6 because it's also positive 1/2 on the y axis and again pi/6 because it's positive 1/2. Since they are less than 270 degrees we would only use pi on the second one to get 30 degrees and nothing on the first solution because it's already a 30 degree angle?
I think you've got it. The only reason I say "I think" is because now I'm the one being confused by what your saying . Anyway, yes for the equation:

$sin(x) = \frac{1}{2}$

we can have $\pi/6$ or $(5\pi)/(6)$, since if you draw a triangle that makes an angle with the x-axis, both of those triangles made make a 30 degree angle with the x-axis, which I think is called the "referance angle", so when your dealing with angles that are greater then $\pi/2$, you can use referance angles and special triangles to determine their correct trigonomic values, and as a result, also figure out different trigonomic equations. Now, does that make sense?

And, no need to apologize for not understanding, everybody has to learn this stuff at some point, and when your first learning it, it doesn't always make perfect sense. Anyway, just post back if you need any more help working out those other problems.

8. yep i get it thank you.

9. Also, I made an error, when I said that you can find soultions to the equations by adding the period, that actually only works if the equation is centered around the x-axis. Which basically means that in this equation:

$y = a\cdot sin(bx+c)+d$

It means that finding other solutions by adding the period only works when:

$d = 0$

In the equation you had, say lets look at:

$sin(x) = -(1/2)$

To graph something like this we need to make it equal to zero (that means the graph hits the x axis there) and then set that expression equal to y. Then, it will give us an equation that, whenever the graph hits the x-axis, is a solution to the original equation:

So we can manipulate the equation to get:

$sin(x) = -(1/2)$

$sin(x) + (1/2) = (1/2) - (1/2)$

$sin(x) + (1/2) = 0$

Setting that equal to y will give us the equation, if you wanted to graph it:

$y = sin(x) + (1/2)$

In this equation $d = 1/2$, so adding the period doesnt work to find new solutions, since the $1/2$ moves the whole graph (the wave looking graph) upward, it no longer its the x-axis is constant intervals; but rather it alternates between a short interval and a large interval. Does that make sense? And if it doesnt I strongly reccomend you graph both $y = sin(x)$ and $y = sin(x) + 1/2$; a picture is worth a thousand words. Anyway, as long as you understand that I made a mistake and it is not always true that you can add the period to find new solutions.