# Solve sec(3θ)−2=0

#### Elissa89

##### Member
Before I post the question I need to vent. I've about had it with my math professor. He isn't showing us how to solve problems that keep popping up in the math homework and I am 100% lost most of the time. Ok I'm done.

The question is

sec 3(theta)-2=0

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Before I post the question I need to vent. I've about had it with my math professor. He isn't showing us how to solve problems that keep popping up in the math homework and I am 100% lost most of the time. Ok I'm done.

The question is

sec 3(theta)-2=0
I'm assuming you mean $\sec^3(\theta)-2=0$?

If so, we can rewrite it as:
$$\sec^3(\theta)=2 \quad\Rightarrow\quad \sec\theta =\sqrt2$$
can't we?

Oh, and can you say what $\sec$ actually is?
Usually we try to express formulas in $\cos$ and $\sin$ after all, which I think makes the analysis a bit easier.

#### Elissa89

##### Member
I'm assuming you mean $\sec^3(\theta)-2=0$?

If so, we can rewrite it as:
$$\sec^3(\theta)=2 \quad\Rightarrow\quad \sec\theta =\sqrt2$$
can't we?

Oh, and can you say what $\sec$ actually is?
Usually we try to express formulas in $\cos$ and $\sin$ after all, which I think makes the analysis a bit easier.
The secant is not cubed, it's sec 3*(theta)

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
Before I post the question I need to vent. I've about had it with my math professor. He isn't showing us how to solve problems that keep popping up in the math homework and I am 100% lost most of the time.
If you have an opportunity, go to the office hour and let the professor know you concerns.

sec 3(theta)-2=0
This can be parsed in many ways: $$\displaystyle \sec(3\theta)-2=0$$, $$\displaystyle \sec(3\theta-2)=0$$, perhaps even $$\displaystyle \sec(3\theta)^{-2}=0$$ or $$\displaystyle s\cdot e\cdot c\cdot 3(\theta)-2=0$$ where $3$ is some function of $\theta$ (why else would you use parentheses around $\theta$ and not the argument of $\sec$?). Also, the problem may ask you to solve the equation, to prove that it has no solutions, to plot the solutions, to prove that solutions are not expressible in radicals or many other things. All this must be in the problem statement.

If you need to solve the equation $\sec(3\theta)-2=0$, then you can proceed as follows.
\displaystyle \begin{align} \sec(3\theta)-2=0&\iff\sec(3\theta)=2\\ &\iff\dfrac{1}{\cos(3\theta)}=2\\ &\iff\cos(3\theta)=\dfrac12\\ &\iff 3\theta=\pm\dfrac\pi3+2\pi k,k\in\mathbb{Z}\\ &\iff\theta=\pm\dfrac\pi9+\dfrac23\pi k,k\in\mathbb{Z} \end{align}

#### Elissa89

##### Member
If you have an opportunity, go to the office hour and let the professor know you concerns.

This can be parsed in many ways: $$\displaystyle \sec(3\theta)-2=0$$, $$\displaystyle \sec(3\theta-2)=0$$, perhaps even $$\displaystyle \sec(3\theta)^{-2}=0$$ or $$\displaystyle s\cdot e\cdot c\cdot 3(\theta)-2=0$$ where $3$ is some function of $\theta$ (why else would you use parentheses around $\theta$ and not the argument of $\sec$?). Also, the problem may ask you to solve the equation, to prove that it has no solutions, to plot the solutions, to prove that solutions are not expressible in radicals or many other things. All this must be in the problem statement.

If you need to solve the equation $\sec(3\theta)-2=0$, then you can proceed as follows.
\displaystyle \begin{align} \sec(3\theta)-2=0&\iff\sec(3\theta)=2\\ &\iff\dfrac{1}{\cos(3\theta)}=2\\ &\iff\cos(3\theta)=\dfrac12\\ &\iff 3\theta=\pm\dfrac\pi3+2\pi k,k\in\mathbb{Z}\\ &\iff\theta=\pm\dfrac\pi9+\dfrac23\pi k,k\in\mathbb{Z} \end{align}
Ummm... thanks but can you fix this so I can read it more easily?

#### MarkFL

Staff member
Ummm... thanks but can you fix this so I can read it more easily?
Are you using a mobile device?

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
I first posted the align environment that was not typeset as LaTeX. I then edited my post and enclosed it it math tags. I am not sure if this is the problem OP referred to.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Are you using a mobile device?
I first posted the align environment that was not typeset as LaTeX. I then edited my post and enclosed it it math tags. I am not sure if this is the problem OP referred to.
Interestingly the problems that we used to have on rendering formulas on mobiles devices seem to have disappeared.
Either way, the formulas that Evgeny posted show up just fine on my mobile device.
I did notice that his original post did not render, but this has been fixed, and the response of the OP actually shows the fixed rendering.

#### MarkFL

Staff member
Interestingly the problems that we used to have on rendering formulas on mobiles devices seem to have disappeared.
Either way, the formulas that Evgeny posted show up just fine on my mobile device.
I did notice that his original post did not render, but this has been fixed, and the response of the OP actually shows the fixed rendering.
It appears that Evgeny's post was quoted after he wrapped his code in math tags. I wasn't aware though that the issue of inline LaTeX bleeding into surrounding text on some mobile devices had gone away. That was the only thing I could think of that would make his otherwise nicely formatted post not be easily read.

#### Greg

##### Perseverance
Staff member
Here's a link to a list of trigonometric identities and related information. It's really quite thorough and contains a nice graphic of the unit circle:

List of trigonometric identities.

#### Elissa89

##### Member
Ummm... thanks but can you fix this so I can read it more easily?
Ok, I get how you solved it, however when I input it the computer tells me it's wrong.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Ok, I get how you solved it, however when I input it the computer tells me it's wrong.
What did you try to input?

#### Elissa89

##### Member
What did you try to input?
Ok I figured it out, I had to include 5*pi/9 +2/3*pi*k

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Ok I figured it out, I had to include 5*pi/9 +2/3*pi*k
For the record, that is only half of the solution.
The full solution is 5*pi/9 +2/3*pi*k (or -pi/9 +2/3*pi*k) combined with pi/9 +2/3*pi*k.

#### Elissa89

##### Member
For the record, that is only half of the solution.
The full solution is 5*pi/9 +2/3*pi*k (or -pi/9 +2/3*pi*k) combined with pi/9 +2/3*pi*k.
I know, I included both

#### Country Boy

##### Well-known member
MHB Math Helper
Before I post the question I need to vent. I've about had it with my math professor. He isn't showing us how to solve problems that keep popping up in the math homework and I am 100% lost most of the time. Ok I'm done.

The question is

sec 3(theta)-2=0
When I was teaching Calculus I had the practice of starting each class by asking if there were any questions about the homework and going over any problems asked about. I also put two or three homework problems on the tests as well as some problems that were just simple variations of those and one or two that were completely different but used the concepts the students should have learned.

One time a student complained bitterly that I had never taught them how to solve this kind of problem. I thought it would be one of those that "were completely different but used the concepts the students should have learned". When I looked at it, it was, in fact, one of the homework problems assigned. It was one that I had gone over in class, and I was able to open the student's notebook and show where he had that specific problem completely solved!

So students (and this was a good student who got a good grade in the course) do not always remember exactly what their teacher has gone over in class!