Usbrth

I am talking about finding a simple, exact, non-trivial solution such as $x = pi/6$. Of course, $x=0$ or $x=pi$ are trivial solutions. I created this equation with its exact solution in mind beforehand, so that's how I know the solution, but if you don't know the solution to begin with, how easy is it to find it, other than using trial and error? Is it an exam question that you could ask to high school students? Another similar equation is $sin x + sin(2x sin x) = 1$ and it also has $x=pi/6$ as a solution. Interestingly, Wolfram Alpha can not find the exact solution, only an approximation (with as many correct digits as you want). See plot of $sin x + sin(2x sin x) - sin 3x$, below.

Just kidding : if you expect analytical solutions for such monsters, stop dreaming !

Now, being serious : take into account that the simple $x=cos(x)$ does not shw analytical solutions and needs to be solved using numerical methods.

Consider that you look for the zero's of
$$f(x)=sin (x) + sin(2x sin (x)) -sin (3x)$$ Plot the function and, visually, locate where (more or less) is the root you want (there is an infinite number of solutions). Call this value $x_0$ and start using Newton method which will update the guess according to
$$x_k+1=x_k-fracf(x_k)f'(x_k)$$

For example, there is root "around" $x=2$. Let us try to get the following iterates
$$left(
beginarraycc
k & x_k \
0 & 2.0000000000000000000 \
1 & 2.2077878077914573503 \
2 & 2.2372992747296569378 \
3 & 2.2389331854770425722 \
4 & 2.2389386369537428867 \
5 & 2.2389386370146724403
endarray
right)$$ which is the solution for twenty significant figures.

The rule here is that anything that appears both inside and outside a transcendental function makes it impossible to solve analytically. Examples are

$$ x - 2 sin(x) = 0 $$
$$ x cos(x)+sin(x) = 1 $$

But you can convert any equation that has sines and cosines of $x$ into a polynomial using the tan-half-angle substitution $t = tan left( fracx2 right)$.

Example, solve:

$$ 3 sin(x)-cos(x) = sqrt6 $$

Use the substitution $x = 2 tan^-1(t)$ in the expression above which yields
$$ beginaligned
sin(x) & = frac2 t1+t^2 \
cos(x) & = frac1-t^21+t^2
endaligned $$

Now solve the expression below in terms of $t$

$$ 3 frac2 t1+t^2 -frac1-t^21+t^2 = sqrt6 $$

$$ t = begincases 1+sqrt6 & text1st solution \
tfrac1+sqrt65 & text2nd solution endcases $$

with the solution in terms of $x$

$$ x = begincases 2 tan^-1left(1+sqrt6right) = 2.57727... & text1st solution \
2 tan^-1left( frac1+sqrt65right) = 1.20783... & text2nd solution endcases $$

Confirmation:

$$ left. 3 sin(2.57727)-cos(2.57727) = sqrt6 phantommatrix\ \ right} 2.4494897...=2.4494897... $$

and

$$ left. 3 sin(1.20783)-cos(1.20783) = sqrt6 phantommatrix\ \ right} 2.4494897...=2.4494897... $$