Trig equations
- Thread starter Loganvega
- Start date
- Joined
- Nov 24, 2012
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Hello, and welcome to FMH! 
The image is sideways, but it looks like:
[MATH]\sec\left(\frac{5}{4}\theta\right)=2[/MATH] where \(0\le\theta<2\pi\)
The first thing I would do, is rewrite the given equation as:
[MATH]\cos\left(\frac{5}{4}\theta\right)=\frac{1}{2}[/MATH]
Now, this leads to the general solution:
[MATH]\frac{5}{4}\theta=2\pi k\pm\frac{\pi}{3}=\frac{\pi}{3}(6k\pm1)[/MATH]
Can you proceed?
The image is sideways, but it looks like:
[MATH]\sec\left(\frac{5}{4}\theta\right)=2[/MATH] where \(0\le\theta<2\pi\)
The first thing I would do, is rewrite the given equation as:
[MATH]\cos\left(\frac{5}{4}\theta\right)=\frac{1}{2}[/MATH]
Now, this leads to the general solution:
[MATH]\frac{5}{4}\theta=2\pi k\pm\frac{\pi}{3}=\frac{\pi}{3}(6k\pm1)[/MATH]
Can you proceed?
- Joined
- Nov 24, 2012
- Messages
- 3,021
To follow up...
[MATH]\theta=\frac{4\pi}{15}(6k\pm1)[/MATH]
For appropriate values of \(k\), we find:
[MATH]\theta\in\left\{\frac{4\pi}{15},\frac{4\pi}{3},\frac{28\pi}{15}\right\}[/MATH]
Here's a plot of the functions:
[MATH]f_1(x)=\sec\left(\frac{5}{4}x\right)[/MATH]
[MATH]f_2(x)=2[/MATH]
And the vertical lines:
[MATH]x_1=\frac{4\pi}{15},\,x_2=\frac{4\pi}{3},\,x_3=\frac{28\pi}{15}[/MATH]
[MATH]\theta=\frac{4\pi}{15}(6k\pm1)[/MATH]
For appropriate values of \(k\), we find:
[MATH]\theta\in\left\{\frac{4\pi}{15},\frac{4\pi}{3},\frac{28\pi}{15}\right\}[/MATH]
Here's a plot of the functions:
[MATH]f_1(x)=\sec\left(\frac{5}{4}x\right)[/MATH]
[MATH]f_2(x)=2[/MATH]
And the vertical lines:
[MATH]x_1=\frac{4\pi}{15},\,x_2=\frac{4\pi}{3},\,x_3=\frac{28\pi}{15}[/MATH]