Thread: it should be easy by now

1. it should be easy by now

Hi folks,

you can see that I have posted quite a few trig questions recently, so I should have it by now!

But sadly no!

say I have cos 2x = 0 and I want the general solution.

$2x = cos^{-1}(0) = 90$ and 270 degrees

so $2x = \dfrac{\pi}{2} + n\pi$ where $n \in \mathcal{Z}$

therefore $x = \dfrac{\pi}{4} + \dfrac{n\pi}{2}$

but this is wrong!

The answer is $n\pi+ \dfrac{\pi}{4}$

in my answer, I put the general term in before dividing by 2 and this seems the way you guys have been doing it in previous posts.
So what's wrong this time?

2. Re: it should be easy by now

I agree with your solution, not "theirs" ...

the cosine function equals zero at odd-multiples of $\dfrac{\pi}{2}$

$\cos(2x)=0 \implies 2x=\dfrac{\pi}{2}(2n+1), \, n\in \mathbb{Z}$

$x=\dfrac{\pi}{4}(2n+1) = \dfrac{\pi}{4}+\dfrac{n\pi}{2}$

so, $x$ is all odd-multiples of $\dfrac{\pi}{4}$

consider $x=\dfrac{3\pi}{4} \implies \cos(2x) = \cos\left(\dfrac{3\pi}{2}\right) =0$

"their" solution skips that value (and other ones) for $x$ ...

$\dfrac{\pi}{4} + n\pi = \dfrac{\pi}{4}, \dfrac{5\pi}{4}, \dfrac{9\pi}{4}, \, ...$

3. Re: it should be easy by now

Your answer is correct. The answer given is only half right, it would need to be \displaystyle \begin{align*} \left\{ \frac{\pi}{4} , \frac{3\,\pi}{4} \right\} + n\,\pi \end{align*}.

4. Re: it should be easy by now

Thanks to skeeter (again) and to Prove It.

I was on the brink of giving up maths altogether!

You have restored my faith.

Thanks Gents

5. Re: it should be easy by now

If you are planning on taking calculus, I encourage you to work overtime and become comfortable with trig. You will encounter a lot of trigonometry in calculus.

6. Re: it should be easy by now

Thanks for the encouragement.