Finding Coordinates for sin(2x)=1/2

[markosheehan]

View attachment 5960
can some one help me with part b finding the co-ordinates of p. i tried this by letting sin 2x=1/2 but when i work out x i do not get the right answer. the right answer is (17pi/12, 1/2)

 

[Greg]

Hi markosheehan. I've re-titled your thread to include a description of the problem.

Here's one way to look at it:

The period of $\sin(2x)$ is $\pi$. One solution is $x=\frac{\pi}{12}$.
The next solution occurs at $x=\frac{\pi}{2}-\frac{\pi}{12}=\frac{5\pi}{12}$.
Since $P$ is one period away from $\frac{5\pi}{12}$, the solution we seek is $(x,y)=\left(\frac{17\pi}{12},\frac12\right)$.

 

[markosheehan]

I still do not understand fully could you explain in a little more detail if possible please

 

[MarkFL]

Another way to look at it is if given:

\(\displaystyle \sin(\theta)=\frac{1}{2}\)

Then by symmetry and periodicity, we see the general solution is:

\(\displaystyle \theta=\frac{\pi}{2}\pm\frac{\pi}{3}+2k\pi=\frac{\pi}{6}(15k\pm2)\) where \(\displaystyle k\in\mathbb{Z}\)

Now, referring to the graph, we see that:

\(\displaystyle \frac{5\pi}{2}<2x<3\pi\)

Now, substitute for $2x$, taking the larger general solution

\(\displaystyle \frac{5\pi}{2}<\frac{\pi}{6}(15k+2)<3\pi\)

\(\displaystyle 15\pi<\pi(15k+2)<18\pi\)

\(\displaystyle 15<15k+2<18\)

\(\displaystyle 13<15k<16\)

\(\displaystyle \frac{13}{15}<k<\frac{16}{15}\)

Hence:

\(\displaystyle k\in\{1\}\)

And so the solution we need is:

\(\displaystyle 2x=\frac{\pi}{6}(15\cdot1+2)\)

\(\displaystyle x=\frac{17\pi}{12}\)

 

[markosheehan]

Sorry I am slow but I don't understand how x can be pi/12 and 5pi/12

 

[MarkFL]

Sorry I am slow but I don't understand how x can be pi/12 and 5pi/12

For my approach, please refer to the following diagram:

View attachment 5961

Now, we should recognize that we must have:

\(\displaystyle \beta=\frac{\pi}{3}\)

And so one solution for \(\displaystyle \sin(\theta)=\frac{1}{2}\) is:

\(\displaystyle \theta=\frac{\pi}{2}\pm\frac{\pi}{3}\)

Now, understanding that the sine function has a period of $2\pi$, we can give the general solution as:

\(\displaystyle \theta=\frac{\pi}{2}\pm\frac{\pi}{3}+2k\pi\) where $k$ is an arbitrary integer.

And we can rewrite this as:

\(\displaystyle \theta=\frac{\pi}{6}(15k\pm2)\)

Does this make sense?