Question

Solve cos 3x = sin 3x; 0 <= x < 2pi
I tried different ways but still couldn't get it.

Answer 1

Hmm here is one approach that might work...

Answer 2

Square each side,\[\Large\bf\sf \cos^23x\quad=\quad \sin^23x\]Subtract sine from each side,\[\Large\bf\sf \cos^23x-\sin^23x\quad=\quad 0\]Apply Cosine Double Angle Identity,\[\Large\bf\sf \cos6x\quad=\quad 0\]

\[\Large\bf\sf \implies\qquad 6x\quad=\quad \frac{\pi}{2}\pm 2k \pi\]

Answer 3

Oh they said 0 to 2pi, my bad.. we don't need all of those rotations.

Answer 4

Err I guess we would want to use quite a few of them, because...
to solve for x, we'll have to divide by 6, right?

So we're going to get a bunch of angles in our interval.

Answer 5

And 3 revolutions will be sufficient, is it? Cause the question itself centers at 3x.

Answer 6

Ah true! :)
Yah that will work I'll bet.

Answer 7

I would have thought we'd include the 3pi/2 angles. But those aren't being included for some reason. Grr I'm too tired -_-

Answer 8

Hmm... This is what I've got for now.
\[x=\frac{ \pi }{ 12 }, \frac{ 5\pi }{ 12 }, \frac{ 7\pi }{ 12 }, \frac{ 11\pi }{ 12 }, \frac{ \pi }{ 4 }, \frac{ 3\pi }{ 4 }\]
The answer that was provided showed that
\[x=\frac{ \pi }{ 12 }, \frac{ 5\pi }{ 12 }, \frac{ 3\pi }{ 4 }, \frac{ 13\pi }{ 12 }, \frac{ 17\pi }{ 12 }, \frac{ 7\pi }{ 4 }\]
I've analysed the answers and I found out that those that are based on the angle \[\frac{ 3\pi }{ 2 }\]
are all wrong.
Any idea what's going on?

Answer 9

My guess is that....
When we squared we introduced some new solutions or something.
So it must have messed things up.


Hmm maybe there was a better way to solve this.
Like dividing by sin3x, and then changing it to tan3x =1 perhaps.. hmmm

Answer 10

If you choose to divide through by sin3x, you get cot3x=1
If you choose to divide through by cos3x, you get tan3x=1

Either method should give us the correct angles.
Ya I think doing the squaring was a bad idea D:

Answer 11

\[\Large\bf\sf \tan3x=1, \qquad\to\qquad 3x=\frac{\pi}{4},\quad \frac{5\pi}{4},\quad \frac{9\pi}{4},\quad ...\]

Answer 12

I think that gets us the correct results, doesn't it? :o

Answer 13

@zepdrix : Well, they are correct.

Answer 14

The same thing happens with polynomials.
Like if you take a simple example like:

x+1 = 2x-1

The solution is clearly x=2.

But if you were to square each side and expand them out,
you end up introducing another solution, x=0.

Squaring can mess things up sometimes D:

Answer 15

I would guess that's what happened here :d
Hmm weird, I thought the division was going to give us trouble..

Answer 16

hmm... True enough. I was also thinking that division would mess things up more for us, and it usually does.

Answer 17

\[
\cos (3x )= \sin (3x)\\
\sin ( \frac \pi 2-3x )= \sin (3x)\\
3x =\frac \pi 2-3x + k \pi \\
3x =\pi -( \frac \pi 2-3x ) + k \pi
\]

Answer 18

One has to consider the first one
\[
3x =\frac \pi 2-3x + k \pi \\
6 x =\frac \pi 2 + k \pi \\
x = \frac {\pi}{12} + \frac { k \pi} 6
\]

Answer 19

For this to be true, sine of one of the corners is equal to cosine of the other corner that isn't the right angle. So we must have a 45-45-90 triangle on our hands.

3x=45 degrees = pi/4 radians

x=15 deg = pi/12 rad

Now the other case is when we rotate another 180 degrees so that both legs are now negative, then the ratio of Opposite over Hypotenuse will be the same as Adjacent over Hypotenuse.

Answer 20

In my post above k should be an even integer

Answer 21

\[

3x =\frac \pi 2-3x + 2 m \,\pi \\
6 x =\frac \pi 2 +2m\, \pi \\
x = \frac {\pi}{12} + \frac { m \,\pi} 3
\]

Answer 22

If it doesn't seem clear how to get to the next point, it's just:

\[3*(\frac{ \pi }{ 12 })+\pi=\frac{ 5 \pi }{ 4 }\]

See how we just plugged in the angle pi/12 to 3x and added pi radians to get us where we wanted?

Answer 23

@eliassaab: Hmm... There's something that I have yet to learn in your solution. Why does when,
\[\sin (\frac {\pi}{2} - 3x) = \sin 3x\]
\[3x = \frac {\pi}{2} - 3x + k\pi\]

Shouldn't that be just
\[3x = \frac {\pi}{2} - 3x\]

How did the k(pi) come by?

Answer 24

That is the final solution
\[
x = \frac {\pi}{12} + \frac { m \,\pi} 3
\]

Answer 25

\[
sin( u+ 2 k \pi)=\sin(u) \\
\]
so if u is a solution them
\[
u+ 2 k \pi
\]
is also a solution

Answer 26

The k is a constant, and pi shows that every 180 degrees you will get to an answer.

So if you look at the picture I drew above, consider:

You need these ones for sin(3x)=cos(3x), otherwise you'll be solving the problem: sin(3x)=-cos(3x) since you can see the 45-45-90 triangle in the second and fourth quadrant have a negative and positive leg each, so they'll be of opposite signs.