MHB Solving Equation 4: Progress & Questions

[Petrus]

Hello MHB,
This is an old exam.What real x satisfies equation $$4\cos^2(x)-4=6\cos(x)$$

Progress:
Subsitute $$u=\cos(x)$$ and I solve this equation
$$4u^2-6u-4=0 $$
$$u_1=-\frac{1}{2}$$ and $$u_2=2$$
so if we take arccos of them we get
$$x=\frac{3\pi}{2}+2k\pi$$ which agree with facit but they got also $$x=-\frac{3\pi}{2}+2k\pi$$ which I don't understand also how shall I know what $$x=\cos^{-1}(2)$$ is in exam? I am doing something wrong or..?

Regards,
$$|\pi\rangle$$

 

[Nono713]

Re: cos equation

Well, $\cos{(x)} = \cos{(x + 2 \pi)}$ by definition ($2 \pi$ is like adding one whole revolution to your angle, so it's the same angle). The inverse cosine function is multivalued, but $\arccos$ is defined as the principal value. Then $\cos{(x)} = \cos{(-x)}$ and so that second value follows (can you see why?)

You need to be careful here because manipulating such multivalued functions can create or destroy solutions to your original equation, so always graph your equation to get a rough idea where the roots are to be sure you didn't miss any and so on.

EDIT: inb4 reply tsunami (Tongueout)

 

[MarkFL]

Re: cos equation

I think you mean to write one of the solutions is:

$$x=\frac{2\pi}{3}+2k\pi$$ where $$k\in\mathbb{Z}$$

and since $\cos(-\theta)=\cos(\theta)$, we also have:

$$x=-\frac{2\pi}{3}+2k\pi$$

which means we may write:

$$x=\frac{2\pi}{3}(6k\pm1)$$

 

[Petrus]

Re: cos equation

Hello MHB,Thanks for fast responed and help! I will have to check this more :) I have indeed seen that $$\cos(x)=\cos(-x)$$ but I don't think I know why but I will think about this and how does it work with sinus?

Regards,
$$|\pi\rangle$$