Solving Equation 4: Progress & Questions

[Petrus]

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

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

Regards,
\(\displaystyle |\pi\rangle\)

 

[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:

\(\displaystyle x=\frac{2\pi}{3}+2k\pi\) where \(\displaystyle k\in\mathbb{Z}\)

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

\(\displaystyle x=-\frac{2\pi}{3}+2k\pi\)

which means we may write:

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

 

[Petrus]

Re: cos equation

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

Regards,
\(\displaystyle |\pi\rangle\)

 

[CellCoree]

Hello |\pi\rangle,

Great job on solving the equation and finding the values of u. The solution x=\frac{3\pi}{2}+2k\pi is correct, but the solution x=-\frac{3\pi}{2}+2k\pi is also valid. This is because the equation has multiple solutions, and both of these values for x satisfy the equation.

As for the value x=\cos^{-1}(2), you are correct in saying that it is not a valid solution. This is because the range of the arccosine function is limited to [-1,1], so any value outside of this range would not be a valid input for the function. In an exam, you should be familiar with the range of the inverse trigonometric functions and be able to identify invalid solutions.

Keep up the good work and continue practicing solving equations to improve your skills.

Best regards,