MHB Find x (Radians): Solve sec(2π/3 + x) = 2

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To solve the equation sec(2π/3 + x) = 2, it can be transformed into cos(2π/3 + x) = 1/2. This leads to the equation 2π/3 + x = kπ ± π/3, where k is an integer. By isolating x, the general solution can be expressed as x = kπ - 2π/3 ± π/3. The discussion highlights the importance of considering the range of the arc-cos function and the periodic nature of cosine when finding solutions. The conversation reflects uncertainty about expanding the solution set appropriately.
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If sec(2pi/3 + x) = 2, what does x equal?

So far I changed it to cos by dividing 1/2. And then, I changed the 1/2 to radians which is pi/3. But, I'm not sure what to do next.
 
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If you are going to find the general solution, you could write:

$$x+\frac{2\pi}{3}=2k\pi\pm\frac{\pi}{3}$$

And then solve for $x$. :D
 
Also you can do it using your way

$$sec(\frac{2\pi}{3}+x)=2$$

$$\frac{1}{cos(\frac{2\pi}{3}+x)}= 2$$

then

$$cos(\frac{2\pi}{3}+x) = \frac{1}{2}$$

$$\frac{2\pi}{3}+x=kx\pi \pm \frac{\pi}{3}$$

solve for x now ..

:)
 
Maged Saeed said:
...
$$cos(\frac{2\pi}{3}+x) = \frac{1}{2}$$

$$\frac{2\pi}{3}+x=kx\pi \pm \frac{\pi}{3}$$

...

How does that follow?
 
MarkFL said:
How does that follow?

$$cos(\frac{2\pi}{3}+x)=\frac{1}{2}$$

$$cos^{-1}cos(\frac{2\pi}{3}+x)=cos^{-1}(\frac{1}{2})$$

$$\frac{2\pi}{3}+x=\frac{\pi}{3}$$

To here , I think I'm correct because the range of arc-cos function is
from zero to PI.
But cannot we expand it to be (2xPI + PI/3) "the solution set of cos function"?

I'm not sure

(Thinking)
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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