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I have a calculus problem that has some trigonometric difficulty to it.

It is $\ 20\sin{x}-10\sin{2x}-\frac{40}{\pi}=0$. I basically want to find two $x \in [0,\pi]$.

I got to $\ \pi\sin{x}(1-\cos{x})=2$

I don't know if there is some trig identity or trick I am missing out on here but I am lost. I found the actual roots from my graphing calculator but would like to know how to go about doing this on my own.

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2 Answers 2

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If $t = \sin(x)$, this can be written as $\pi t (1\pm\sqrt{1-t^2}) = 2$, or $\pm\sqrt{1-t^2} = 2/(\pi t) - 1$. Squaring both sides and simplifying, we get the quartic $$ \pi^2 t^4-4 \pi t+4 = 0$$ which has two real roots, approximately $.3273258251$ and $.9452058404$. There are formulas for solving the quartic in radicals, but they aren't terribly pretty.

For $t = .3273258251$ we want $\cos(x) < 0$, thus $x = \pi - \arcsin(t) \approx 2.808120551$, while for $t = .9452058404$ we want $\cos(x) > 0$, thus $x = \arcsin(.9452058404) \approx 1.238224522$.

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  • $\begingroup$ what's a quartic? And how do I know to use $\pm\sqrt{1-t^2}$? $\endgroup$
    – Computer
    Commented Jun 29, 2017 at 1:23
  • $\begingroup$ A quartic is a polynomial of degree 4. $\endgroup$ Commented Jun 29, 2017 at 4:08
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Similar to Robert Israel's answer.

Using the tangent half-angle substitution $$t=\tan(\frac x 2)\qquad \sin(x)=\frac{2t}{1+t^2}\qquad \cos(x)=\frac{1-t^2}{1+t^2}$$ the equation reduces to $$t^4-2 \pi t^3+2 t^2+1=0$$ Just as Robert Israel answered, solving analytically quartic equations is not the most pleasant thing to do and numerical methods are required.

Looking here, the discriminant is given by $\Delta=16 \pi ^2 \left(64-27 \pi ^2\right) <0$ and then the equation has two distinct real roots and two complex conjugate non-real roots.

Looking at the plot of the function, we can see that one root is "close" to $1$ and the second "close" to $6$. So, let us use Newton method to solve for the roots. The successive iterates will be $$\left( \begin{array}{cc} n & t_n \\ 0 & 1.000000 \\ 1 & 0.789560 \\ 2 & 0.720530 \\ 3 & 0.712669 \\ 4 & 0.712570 \end{array} \right)$$ $$\left( \begin{array}{cc} n & t_n \\ 0 & 6.00000 \\ 1 & 5.94350 \\ 2 & 5.94182 \end{array} \right)$$

which are the solutions for six significant figures.

So, $$t_1=0.712570 \implies \frac {x_1}2=\tan^{-1}(0.712570)\implies x_1=1.23822$$ $$t_2=5.94182 \implies \frac {x_2}2=\tan^{-1}(5.94182)\implies x_2=2.80812$$

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