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

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To find the coordinates for the equation sin(2x) = 1/2, the period of sin(2x) is π, leading to initial solutions at x = π/12 and x = 5π/12. The correct coordinate is identified as (17π/12, 1/2) by considering the periodic nature of the sine function and the general solution for sin(θ) = 1/2. The discussion emphasizes the importance of understanding both the symmetry and periodicity of the sine function to derive the correct values. Clarifications were requested regarding the multiple solutions, highlighting the complexity of the problem.
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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)
 

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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 $(x,y)=\left(\frac{17\pi}{12},\frac12\right)$.
 
I still do not understand fully could you explain in a little more detail if possible please
 
Another way to look at it is if given:

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

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

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

Now, referring to the graph, we see that:

$$\frac{5\pi}{2}<2x<3\pi$$

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

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

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

$$15<15k+2<18$$

$$13<15k<16$$

$$\frac{13}{15}<k<\frac{16}{15}$$

Hence:

$$k\in\{1\}$$

And so the solution we need is:

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

$$x=\frac{17\pi}{12}$$
 
Sorry I am slow but I don't understand how x can be pi/12 and 5pi/12
 
markosheehan said:
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:

$$\beta=\frac{\pi}{3}$$

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

$$\theta=\frac{\pi}{2}\pm\frac{\pi}{3}$$

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

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

And we can rewrite this as:

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

Does this make sense?
 

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