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

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    2017
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High School POTW #289 features a trigonometric and polynomial equation for members to solve. The equation is \(\frac{1}{4}\left(\sin\left(\frac{\pi x}{2}\right)\right)^2+2x^4-5x^2+1=0\). The organizer acknowledges a delay in posting and will provide two problems this week to make up for it. Opalg is recognized for submitting the only complete solution, while MarkFL and lfdahl received partial credit for their correct but incomplete solutions. Members are encouraged to participate and follow the provided guidelines for submissions.
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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:

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

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