Trigonometry Question

You should solve this problem by trying things that you think my work. Exactly what have you tried? Where are you getting stuck? Maybe you should try to write the equation so that all the angles are the same.
Please show us your attempts so we have an idea on how you want to solve this problem and we can see where you are getting stuck.
 
I was trying to transform everything either to cos(x) or sin(x).
I was working as follows:
sin(3x)+sin(7x)=2
sin(3x)+sin(3x+4x)=2
sin(3x)+sin(3x)cos(4x)+cos(3x)sin(4x)=2

I don't know if I am on the right way.
 
nicoleoliva said:
I was trying to transform everything either to cos(x) or sin(x).
I was working as follows:
sin(3x)+sin(7x)=2
sin(3x)+sin(3x+4x)=2
sin(3x)+sin(3x)cos(4x)+cos(3x)sin(4x)=2

I don't know if I am on the right way.
Click to expand...

Where did that 4x in the second step come from? Where is the rest of the problem?
 
Yep, after looking at your equation again, that 2 really stuck out!
Since -1< sin(x) < 1 for ALL x values, the only way sin(mx) + sin(nx) = 2 is IFF sin(mx)=sin(nx) = 1
MarkFL saw this right away I suspect. I will be in the corner thinking about how I missed that one!
 
topsquark said:
nicoleoliva was splitting the 7x into 3x + 4x to use the sum of angles formula. Not a bad approach.

-Dan
Click to expand...
I respectfully disagree with you. With that sum being 2, there is only one way that can happen and that is if both those sin function equaling 1.
 
Jomo said:
I respectfully disagree with you. With that sum being 2, there is only one way that can happen and that is if both those sin function equaling 1.
Click to expand...
Agreed. And I admit that I missed that entirely. But if we didn't have that information then I would take nicoleoliva's attempt a good try.

-Dan
 
topsquark said:
And I admit that I missed that entirely.
Click to expand...
Dan. I too missed that 2 initially. However after I realized that 2 was the key to solving this problem I went to the corner for 2 hours to think about it.

Are you going to the corner?
 
I graphed the two functions separately. Is that cheating?

We need values of x where both functions reach their maximum (1) at the same time.
 
bZNyQ7C2 said:
I graphed the two functions separately. Is that cheating?

We need values of x where both functions reach their maximum (1) at the same time.
Click to expand...
No - that is not cheating. It is fair use of alternative resources (unless explicitly forbidden).

So what was (were) the answer/s?
 
Well, there's a point at -1/2 pi and one at 3/2 pi. The period of 2 pi.

\(\displaystyle - \frac{1}{2}\pi + 2k\pi \quad k \in Z\)

There's still something about the algebra that isn't clear in my mind yet.

How do you get that double-stroke Z in Latex?
 
Jomo said:
Dan. I too missed that 2 initially. However after I realized that 2 was the key to solving this problem I went to the corner for 2 hours to think about it.

Are you going to the corner?
Click to expand...
Only if we can go there together. I get lonely in the corner.

-Dan
 
bZNyQ7C2 said:
Well, there's a point at -1/2 pi and one at 3/2 pi. The period of 2 pi.

\(\displaystyle - \frac{1}{2}\pi + 2k\pi \quad k \in Z\)

There's still something about the algebra that isn't clear in my mind yet.

How do you get that double-stroke Z in Latex?
Click to expand...
you figured out roots of:

f(x) = sin(3x) + sin(7x) - 2

We know that both sin(3x) and sin(7x) will be equal to 1 (our criterion) at x = -\(\displaystyle \frac{\pi}{2}\)
 
You can figure this out by algebra and a simple Diophantine equation.

[MATH]sin(3x) = 1 \implies 3x = \dfrac{\pi}{2} + 2k\pi \implies[/MATH]
[MATH] \dfrac{x}{\pi} = \dfrac{1}{6} + \dfrac{2k}{3} = \dfrac{1}{6} * (1 + 4k).[/MATH] And

[MATH]sin(7x) = 1 \implies 7x = \dfrac{\pi}{2} + 2j \pi \implies[/MATH]
[MATH] \dfrac{x}{\pi} = \dfrac{1}{14} + \dfrac{2j}{7} = \dfrac{1}{14} * (1 + 4j).[/MATH]
Two things equal to a third equal each other.

[MATH]\therefore \dfrac{1}{6} * (1 + 4k) = \dfrac{1}{14} * (1 + 4j) \implies 14(1 + 4k) = 6(1 + 4j) \implies[/MATH]
[MATH]14 + 56k = 6 + 24j \implies 8 = 24j - 56k \implies 1 = 3j - 7k.[/MATH]
But j and k must be integers. k = 2 and j = 5 are the smallest positive integers that solve that equation.

Therefore the smallest positive x that solves the original equation is

[MATH]\dfrac{x}{\pi} = \dfrac{1}{6} * (1 + 4 * 2) = \dfrac{3}{2}\implies x = \dfrac{3\pi}{2}.[/MATH]
Let's check.

[MATH]sin \left ( 3 * \dfrac{3\pi}{2} \right ) = sin \left ( \dfrac{9\pi}{2} \right ) = sin \left ( \dfrac{\pi}{2} + 4\pi \right ) = sin \left ( \dfrac{\pi}{2} + 2 * 2 \pi \right ) = 1.[/MATH]
[MATH]sin \left ( 7 * \dfrac{3\pi}{2} \right ) = sin \left ( \dfrac{21\pi}{2} \right ) = sin \left ( \dfrac{\pi}{2} + 10\pi \right ) = sin \left ( \dfrac{\pi}{2} + 2 * 5 \pi \right ) = 1.[/MATH]
Note that this is equivalent to your graphing solution because

[MATH]\dfrac{3\pi}{2} + 2(-\ 1)\pi = - \dfrac{\pi}{2}.[/MATH]
Of course, this is all dependent on Mark's insight. In theory, you should be able to do it by using the angle sum formula, but I guarantee I would get lost in that mess somewhere.
 
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