# How to solve for $x$ in $x(x^3+\sin x \cos x)-\sin^2 x =0$?

How do I solve for $x$ in $$x\left(x^3+\sin(x)\cos(x)\right)-\big(\sin(x)\big)^2=0$$

I hate when I find something that looks simple, that I should know how to do, but it holds me up.

I could come up with an approximate answer using Taylor's, but how do I solve this?

(btw, WolframAlpha tells me the answer, but I want to know how it's solved.)

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Maybe some wise inequality does a good job here. – user 1618033 Aug 22 '12 at 16:57

We prove that $x=0$ is the only solution.

Let

$$f(x)= x^4+x \sin(x) \cos(x)- \sin^2(x) \,.$$

Then $f$ is even, so it is enough to look for roots on $[0, \infty)$.

You cana lso observe that $f(x)\geq x^4 -x-1$, and an easy calculation shows that for all $x> \sqrt[3]{2}$ we have $x^4-x-1 >0$.

Thus the only possible positive roots are in the interval $[0, \sqrt[3]{2}]$, which is inside the first quadrant.

Then for all $x \neq 0$, by using $x >\sin(x)$ we have

$$x^4+x\sin(x)\cos(x)-\sin^2(x) >x^2 \sin^2(x)+\sin^2(x)\cos(x)-\sin^2(x)$$ $$=\sin^2(x)(x^2+\cos(x)-1)>\sin^2(x)(x^2+\cos^2(x)-1)=\sin^2(x)(x^2-\sin^2(x))>0$$

P.S. Just to clarify, we need to make first the reduction to the first quadrant to make sure than $x> \sin(x)$ implies inequalities of the type $x\sin(x)\cos(x) > \sin^2(x)\cos(x)$

P.P.S I also suspect that $f'(x) >0$ for $x>0$, which would lead to a second proof of the problem. Note that $$f'(x)=4x^3+x \cos(2x)-\frac{1}{2}\sin(2x)$$ which can easly be proven to be positive on $(0,\frac{\pi}{2}]$ and $(\frac{\pi}2, \infty)$.

$$f''(x)= 12x^2 -2x \sin(2x)$$

If $x>0$ then since $\sin(2x)<2x$ we have

$$12x^2 -2x \sin(2x)> 12x^2-2x\cdot(2x)>0$$

Thus $f'(x)$ is strictly increasing on $[0, \infty)$. Since $f'(0)=0$ we get that $f'(x)>0$ for all $x>0$. Thus $f(x)$ is strictly increasing on $[0, \infty)$, and since $f(0)=0$, it follows that $f(x)=0$ has no solution on $(0, \infty)$. Since $f(x)$ is even, it follows that $x=0$ is the unique solution.

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Only solution, or merely only real solution? – GEdgar Aug 22 '12 at 16:25
Good point, only one real solution. – N. S. Aug 22 '12 at 16:35
@N. S.: for the second solution you need no more than the 3rd row, but a bit modified. – user 1618033 Aug 23 '12 at 7:49
And, this second solution is probably the intended solution, but it's hard to say without context. – Graphth Aug 24 '12 at 15:28

Polynomials and trig functions don't play nice together, so you are usually stuck with numeric solutions. You can start by noting that $x=0$ is a double root, one from the outer $x$ and one from the $x/\sin x$ terms.

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I got that the equation only has the root $x=0$, it seems to be an easy "unique solution" equation.. Unless I made a mistake.... – N. S. Aug 22 '12 at 15:59
@N.S.: I misread the problem and thought the outside $x$ multiplied the whole thing. I stripped it off and there were two more roots. Solve a different problem, get a different answer. I like your solution. – Ross Millikan Aug 22 '12 at 16:21

Let's do it in a simple way. As N. S. noticed, the function is even. Then it's enough to analyze things on the positive real axis:

$$x\left(x^3+\sin(x)\cos(x)\right)-\big(\sin(x)\big)^2\geq x\left(x^3+x-x^3\right)-\big(\sin(x)\big)^2\geq 0$$ $$x^2 \geq \big(\sin(x)\big)^2$$

Above I used the fact that $\sin(x)\cos(x)\geq x-x^3$ for $x \ge 0$ for that we may use the following proof (or enter here for more nice proofs):

Let's consider

$$f(x) = \sin(x) \cos(x)-x+x^3$$ then $$f'(x) = 3 x^2-2\sin^2(x)\tag1$$ $$x\ge \sin(x)\tag2$$ From $(1)$ and $(2)$ we immediately notice that $f'(x)\ge0$ and taking into account that $f(0)=0$ we may conclude that the inequality holds. The equality is obviously reached only when $x=0$.
Hence the only solution is got for $x=0$.

Q.E.D.

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Using the identity $\cos x=1-2\sin^2(x/2)$ and introduccing the function ${\rm sinc}(x):={\sin x\over x}$ we can rewrite the given function $f$ in the following way: $$f(x)=x^2\left(x^2\left(1-{1\over2}{\rm sinc}(x){\rm sinc}^2(x/2)\right)+{\rm sinc}(x)\bigl(1-{\rm sinc}(x)\bigr)\right)\ .\qquad(*)$$ Now ${\rm sinc}(x)$ is $\geq0$ on $[0,\pi]$ and of absolute value $\leq1$ throughout. By distinguishing the cases $0<x\leq\pi$ and $x>\pi$ it can be verified by inspection that $f(x)>0$ for $x>0$. Since $f$ is even it follows that $x_0=0$ is the only real zero of $f$.

[One arrives at the representation $(*)$ by expanding the simple functions appearing in the given $f$ into a Taylor series around $0$ and grouping terms of the same order skillfully.]

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