Can You Solve High School POTW #289's Trigonometric and Polynomial Equation?

[anemone]

Hi MHB,

The High School POTW should be number 289 this week, but due to the fact that I was a bit late carrying out my duty over several weeks, I fell behind a week, and so I will make it up by posting two POTWs this week.

I sincerely apologize for this, and I hope our members can take up the challenge and solve two High School POTWs this week!

Here is this week's second POTW:

-----

Solve \(\displaystyle \frac{1}{4}\left(\sin\left(\frac{\pi x}{2}\right)\right)^2+2x^4-5x^2+1=0.\)

-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to Opalg for submitting the only solution that can truly be considered complete. His solution is shown below:

Spoiler

Using the trig formula $\sin^2\theta = \frac12(1 - \cos(2\theta)$, the equation becomes $\frac18(1 - \cos(\pi x)) + 2x^4 - 5x^2 + 1 = 0.$ Write that as $$0 = 16x^4 - 40x^2 + 9 - \cos(\pi x) = (4x^2 - 1)(4x^2 - 9) - \cos(\pi x).$$

Let $f(x) = (4x^2 - 1)(4x^2 - 9)$ and $g(x) = \cos(\pi x).$ Then $f(x) = 0$ when $x = \pm\frac12$ and when $x = \pm\frac32$. But $g(x)$ is also zero at each of those four points. So that gives four solutions to the original problem $f(x) - g(x) = 0.$

The fact that these are the only solutions is "obvious" from the graphs of $f(x)$ and $g(x)$, but it seems harder to give a formal proof.

[DESMOS=-2.4417637271214643,2.5582362728785357,-2.3125000000000373,2.6874999999999627]y\ =\ \left(4x^2-1\right)\left(4x^2-9\right);y\ =\ \cos\left(\pi x\right)[/DESMOS]

Both functions $f(x)$ and $g(x)$ are even, so it is enough to show that $\frac12$ and $\frac32$ are the only positive solutions. The derivative $f'(x) - g'(x)$ is $16x(4x^2-5) + \pi\sin(\pi x)$. This is negative for $0<x<1$, so $f(x) - g(x)$ can only have one zero in that interval. The derivative is positive whenever $x>\frac54$, so there can only be one zero in that interval. Finally, in the interval $1\leqslant x\leqslant \frac54$ the function $f(x)$ is less than $-14$, so that $f(x) - g(x)$ is negative and cannot have any zeros.

The following members found the correct solutions, but did not demonstrate that they were the only solutions, and so they get partial credit:
1. MarkFL
2. lfdahl