Solving sin3x=1/2: 6 Solutions Explained

[seiferseph]

sin3x=1/2

how would i go about solving this problem? and there should be 6 solutions right, how do i get them? thanks.

 

[ToxicBug]

The answer is x = pi/18 but I don't know how to really get it. I just thought what sin x = 1/2 (x = pi/6) and since its sin 3x, then its that /3 (pi/18).

 

[BobG]

There's six solutions because:

a) there's two angles between 0 and 2 \pi that have a sine of 1/2

b) you have 3x equals one of those two angles. In other words, for each of the angles that have a sine of 1/2, there's three angles between 0 and 2 \pi that, if multiplied by 3, would wind up at the same spot on the unit circle.

Example: \frac{13 \pi}{18} * 3 = \frac{13 \pi}{6} which is the same spot on the unit circle as \frac{\pi}{6}

Find each of the angles that have a sine of 1/2

Divide them by 3.

Add 2 \pi to the original angles, then divide by 3 again.

Add 4 \pi to the original angles, then divide by 3 again.

All three should lie between 0 and 2 pi.

If you added 6 \pi to the original angles and divided by 3, you would find that your answer was greater than 2 \pi (should be obvious, since 6 \pi divided by 3 is 2 \pi).

 

[dextercioby]

\sin 3x=\sin \ (x+2x)=\sin x \ (\cos^{2}x-\sin^{2}x)+\cos x \ (2\sin x\cos x)=...=-4\sin^{3}x+3\sin x...

The u have to solve a cubic,assuming:
\sin x=u

-4u^{3}+3u-\frac{1}{2}=0...

Daniel.

 

[BobG]

\sin 3x=\sin \ (x+2x)=\sin x \ (\cos^{2}x-\sin^{2}x)+\cos x \ (2\sin x\cos x)=...=-4\sin^{3}x+3\sin x...

The u have to solve a cubic,assuming:
\sin x=u

-4u^{3}+3u-\frac{1}{2}=0...

Daniel.

Wow! That's an interesting solution. Are you solving the same problem he's asking? (He never did actually state what he was solving for. I just kind of assumed he was trying to solve for x.)

Edit: But, your solution does give the sine of x. A point worth noting, since most people only memorize the basic angles. If you want the sine of pi/18, you would have to multiply your angle by 3 to get an angle you did know the sine of and then solve for sine x the way dex did.

 

[dextercioby]

Bob,you're not funny.

Daniel.

 

[seiferseph]

didn't have a chance to try anything yet (thanks so much for the help!), but its solving for x, and therefore all the possible angles

 

[seiferseph]

thanks, i think i got that one. now for cox3x=1/2, this is what i did (is it right?)
ok, so i put 3x = pi/3, 5pi/3. then divide each by 3 to get
x=pi/9, 5pi/9
add 2pi/3 and 4pi/3

for the solutions i get (Π is pi)
x=Π/9, 5Π/9, 7Π/9, 11Π/9, 13Π/9, 17Π/9

are these correct for the 6 solutions?

 

[dextercioby]

thanks, i think i got that one. now for cox2x=1/2, this is what i did (is it right?)
ok, so i put 3x = pi/3, 5pi/3. then divide each by 3 to get
x=pi/9, 5pi/9
add 2pi/3 and 4pi/3

for the solutions i get (Π is pi)
x=Π/9, 5Π/9, 7Π/9, 11Π/9, 13Π/9, 17Π/9

are these correct for the 6 solutions?

That is INCOMPATIBLE with the calcuations that u made...

Daniel.

 

[p53ud0 dr34m5]

you have 2x this time. using 3x will not help you.

 

[seiferseph]

sorry, it was actually cos3x=1/2. can anyone confirm those solutions?

 

[futb0l]

\cos{3x} = \frac{1}{2}

So...
remember that cos is positive in first and fourth quadrant...

<br /> 3x = \frac{\pi}{3}, -\frac{\pi}{3}, \frac{5\pi}{3} ...<br />

<br /> x = \frac{\pi}{9}, -\frac{\pi}{9}, \frac{5\pi}{9} ...<br />

Is there a restriction on the angles?? Because you can just keep on adding two pi onto them and it will be a valid answer.

 

[seiferseph]

\cos{3x} = \frac{1}{2}

So...
remember that cos is positive in first and fourth quadrant...

<br /> 3x = \frac{\pi}{3}, -\frac{\pi}{3}, \frac{5\pi}{3} ...<br />

<br /> x = \frac{\pi}{9}, -\frac{\pi}{9}, \frac{5\pi}{9} ...<br />

Is there a restriction on the angles?? Because you can just keep on adding two pi onto them and it will be a valid answer.

0 ≤ x < 2pi
is the restriction

(and you keep adding 2pi/3, or the period, right?)

 

[futb0l]

Oopps, this is supposed to be the answers:

<br /> 3x = \frac{\pi}{3}, \frac{5\pi}{3} , \frac{7\pi}{3} , \frac{11\pi}{3} , \frac{13\pi}{3}, \frac{17\pi}{3}, \frac{19\pi}{3} ...

Because the its in the 1st quadrant and the 4th quadrant, it is pi/3 and 5pi/3, because it is in a circle, you can just keep adding 2pi (360 degrees) to pi/3 or 5pi/3... but when you divide by 3, it has to be within 0 < x < 2pi

remember that the cos function has a period of 2pi, so it's a repeating cycle -- think of the cos graphs.

so... u divide each answer by 3...

<br /> x = \frac{\pi}{9}, \frac{5\pi}{9} , \frac{7\pi}{9} , \frac{11\pi}{9} , \frac{13\pi}{9}, \frac{17\pi}{9}<br />

As you can see, I didnt include 19pi/9 as an answer because it is outside the domain.

 

[Curious3141]

sin3x=1/2

how would i go about solving this problem? and there should be 6 solutions right, how do i get them? thanks.

The thing to remember is that for trig equations, there are potentially an *infinite* number of solutions. This is because trig functions are periodic.

So you must always specify the range you're looking for solutions in. This is usually 0 \leq x &lt; 2\pi but not always. So this must be specified.

Assuming that's the range we're going to use, this is the simplest way to solve it :

Think about what angles have sines of half. For sine to be positive, the angle can be in the 1st and 2nd quadrants, so you have \frac{\pi}{6} and \frac{5\pi}{6}. Those are two possible values for 3x, correct ?

Since you're going to be dividing those by 3 to get x, you should take more values for 3x if you want to get all the values in the required range. Since you know \sin \theta = \sin{(\theta + 2k\pi)}, where k is an integer, just add 2\pi and 4\pi to the above values to get all the possible values of 3x.

So you have :

3x = \frac{\pi}{6}, \frac{5\pi}{6}, (\frac{\pi}{6} + 2\pi), (\frac{5\pi}{6} + 2\pi), (\frac{\pi}{6} + 4\pi), (\frac{5\pi}{6} + 4\pi)

and dividing by 3 and simplifying, you get the required solutions in radians within the required range.

 

[seiferseph]

ok, thanks to everyone, i got the answers (which i probably could've done) but more importantly know how to do it and why it works. thanks!