# Solving Within an Interval & Finding the Period

• Aug 24th 2013, 04:32 PM
Fratricide
Solving Within an Interval & Finding the Period
1) Find the solutions of http://i.imgur.com/BtdJaZQ.pngin the interval http://i.imgur.com/FH51U7A.png

I can determine that there are 4 solutions that need to be found, but I can only find 2 of them (http://i.imgur.com/yd5UOCc.png). How do I find the others?

2) Find the period of the graph http://i.imgur.com/CBshEEa.png

I know how to find the period of functions containing either sin or cos, but not both. Can someone give me a crash course on how to do this?

(Apologies for the choppy equation structure.)
• Aug 24th 2013, 04:56 PM
MathoMan
Re: Solving Within an Interval & Finding the Period
$\displaystyle 2\sin 3x +\sqrt{2}=0$
$\displaystyle \sin 3x=-\frac{\sqrt{2}}{2}$

Draw a unit trigonometric circle, find the value $\displaystyle -\frac{\sqrt{2}}{2}$ on y-axis, and . Draw a line parallel to x-axis that runs trough $\displaystyle -\frac{\sqrt{2}}{2}$ on the y-axis. The line intersects the unit circle in two distinct points - numbers that are at those two points on the trigonometric circle are the solutions to the given equation.

All of the solutions can be found solving $\displaystyle 3x=\frac{5\pi}{4}+k\cdot 2\pi$ and $\displaystyle 3x=\frac{7\pi}{4}+k\cdot 2\pi$, $\displaystyle k\in \mathbf{Z}$.
• Aug 24th 2013, 05:10 PM
Shakarri
Re: Solving Within an Interval & Finding the Period
1) Remember that $\displaystyle sin(3x)=sin(3x+2\pi)=sin(3x+4\pi)$ When you find a solution to the equation you can add 2*pi to it again and again to get more solutions but you have an upper limit so the number of solutions will be limited.

2) The period P will satisfy the equation f(x)=f(x+P)
• Aug 24th 2013, 05:59 PM
Shakarri
Re: Solving Within an Interval & Finding the Period
I realise that my suggestion for part 2) is very difficult if not impossible to solve.
You cannot guarantee this will be the smallest value of p to satisfy the equation but the function will repeat if there is a value y so that $\displaystyle 4sin(3\pi x)=4sin(3\pi(x+y))$ and $\displaystyle -3cos(2\pi x)=-3cos(2\pi(x+y))$

Remember that $\displaystyle sin(x+2\pi n)=sin(x)$
same for cos
y will have to be an integer
• Aug 25th 2013, 10:01 PM
Fratricide
Re: Solving Within an Interval & Finding the Period
Quote:

Originally Posted by MathoMan

All of the solutions can be found solving $\displaystyle 3x=\frac{5\pi}{4}+k\cdot 2\pi$ and $\displaystyle 3x=\frac{7\pi}{4}+k\cdot 2\pi$, $\displaystyle k\in \mathbf{Z}$.

I'm not quite sure I fully understand. What's the value of k?