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Solve for solutions from $0\le x<2\pi$:

$$\sec(4x)=2$$

When I solve this equation my answer comes to be $x=\pi/12,5\pi/12$.

However when I graph the equation $y=\sec(4x)-2$, there 6 other values for $x$ that equal zero.

How would I find these values for $x$ using the previous answers I found?

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  • $\begingroup$ it is only $$x=\frac{\pi}{3}$$ $\endgroup$ Jul 16, 2017 at 18:21
  • $\begingroup$ If you plug π/3 into sec(4x) it would equal -2 not 2. Also there are more than one possible values for x that would result in sec(4x) equaling 2 $\endgroup$
    – Ludwig
    Jul 16, 2017 at 18:25
  • $\begingroup$ oh sorry i wrote $$\sec(x)=2$$ instead $$\sec(4x)=2$$ $\endgroup$ Jul 16, 2017 at 18:28
  • $\begingroup$ Also I originally wrote the domain for the solutions of x incorrectly, it's 0 ≤ x < 2π $\endgroup$
    – Ludwig
    Jul 16, 2017 at 18:30
  • $\begingroup$ it is really $$x=\frac{\pi}{12},\frac{5\pi}{12},\frac{7\pi}{12},\frac{11\pi}{12}$$ $\endgroup$ Jul 16, 2017 at 18:33

2 Answers 2

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This is a very common mistake that I see a lot among students — you applied the given constraint inappropriately. I bet your reasoning was as follows:

$\sec(4x)=2$ has only two solutions within the interval $[0,2\pi)$, which are $\pi/3$ and $5\pi/3$. So $4x=\pi/3,5\pi/3$, and dividing by $4$ we get $x=\pi/12,5\pi/12$.

Your mistake is that you effectively said that $\color{red}{4x\text{ is in } [0,2\pi)}$, but that is NOT the given requirement — the given constraint is that $\color{green}{x\text{ is in } [0,2\pi)}$. Again: in your first step you were solving for $4x$, and you chose values for $4x$ only within $[0,2\pi)$, but $4x$ is NOT required to lie within this interval. You should only apply this constraint to values of $x$, not of anything else.

Here's how I suggest solving constrained trig equations. First of all, forget about the constraint and find ALL solutions:

$$\sec(4x)=2 \quad \Longrightarrow \quad 4x=\frac{\pi}{3}+2\pi n,\frac{5\pi}{3}+2\pi n \quad \Longrightarrow \quad x=\frac{\pi}{12}+\frac{\pi n}{2},\frac{5\pi}{12}+\frac{\pi n}{2}.$$

Now, that you have the values of $x$ (not of something else, like $4x$), it's time to check which ones of them lie withing the required interval. This way you will find all requested solutions to the equation. In this example, in both families $n=0,1,2,3$ produce values of $x$ within $0\le x<2\pi$, so there are $8$ solutions.

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Well, note that we have the following:

  • $\sec(4x)=2$ if and only if $\cos(4x)=\frac12$
  • $0\leq x<2\pi$ if and only if $0\leq 4x<8\pi$

Thus, you need only find the solutions of $\cos(t)=\frac12$ for $0\leq t<8\pi,$ and convert to solutions of the given equation via $x=\frac14t.$

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