$How do you go about solving this question?$

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$How do you go about solving this question?$

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The answers are $-\frac{5\pi }{24} ,-\frac{\pi }{24},\frac{19\pi }{24},\frac{23\pi }{24} $ But I don't know how you get $\frac{19\pi }{24},\frac{23\pi }{24} $ whenever I try and solve it I only get $-\frac{5\pi }{24} ,-\frac{\pi }{24}$.

Trigonometry Algebra

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$\dfrac{1}{2}\sec\left(2\phi+\dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt3}$

$\sec\left(2\phi+\dfrac{\pi}{4}\right) = \dfrac{2}{\sqrt3}$

$\cos\left(2\phi+\dfrac{\pi}{4}\right) = \dfrac{\sqrt3}{2}$

This means that $2\phi+\dfrac{\pi}{4} = \dfrac{\pi}{6} + 2k\pi$ or $2\phi+\dfrac{\pi}{4} = -\dfrac{\pi}{6} + 2k\pi$, $k \in \mathbb{Z}$ (remember that cosine is an even periodic function).

which means that $\phi = -\dfrac{\pi}{24} + k\pi$ or $\phi = -\dfrac{5\pi}{24} + k\pi, k \in \mathbb{Z}$

Now, since we need $-\pi < \phi < \pi$, by inspection we can see that the solutions must be:

$\phi = -\dfrac{\pi}{24}, -\dfrac{\pi}{24}+ \pi, -\dfrac{5\pi}{24}, -\dfrac{5\pi}{24} + \pi$, which coincides with the given answers.

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Where do we get the 2kπ from?
- Mathe
  
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  Cosine is periodic with period 2π. This means that cos(x) = cos( x + 2π ) = cos(x - 2π) = cos(x+4π) = cos(x-4π). So if x is a solution to cos(x) = sqrt(3)/2, x+2kπ will also be a solution,

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