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)
 
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