Resolve: $4\sin(2x)+4\cos(x)-5=0$ The first thing that comes to mind is to substitute $\sin(2x)=2\sin(x)\cos(x)$ and so we have:
\begin{align*}
8\sin(x)\cos(x)+4\cos(x)-5=0
\end{align*}
But after that I can't see what other identity to apply, I've been checking several times, if it is necessary to add something or multiply properly, but I can't find anything.
 A: HINT
By tangent half angle identities we obtain
$$\frac{16t(1-t^2)}{(1+t^2)^2}+\frac{4(1-t^2)}{1+t^2}-5=0$$
$$\iff 9 t^4 + 16 t^3 + 10 t^2 - 16 t +1=0$$
which can be studied by IVT to show that two real solutions exist.
A: If you plot the function
$$f(x)=4\sin(2x)+4\cos(x)-5$$ by inspection or drawing, you will notice that, for $0 \leq x \leq 2\pi$, there are two roots; one of them is close to $0$ and the other one close to $\frac \pi 3$.
Using Taylor expansions, we have
$$f(x)=-1+8 x-2 x^2-\frac{16 x^3}{3}+\frac{x^4}{6}+O\left(x^5\right)$$ Using now series reversion
$$x=t+\frac{t^2}{4}+\frac{19 t^3}{24}+\frac{57 t^4}{64}+O\left(t^5\right)\qquad \text{with} \qquad t=\frac {1+f(x)}8$$ Making $f(x)=0$ as desired, an estimate is
$$x=\frac{102763}{786432}=0.130670$$ while the solution given by Newton method is $x=0.130753$.
Doing the same around $x=\frac \pi 3$
$$f(x)=\left(2 \sqrt{3}-3\right)-\left(4+2 \sqrt{3}\right) \left(x-\frac{\pi
   }{3}\right)-\left(1+4 \sqrt{3}\right) \left(x-\frac{\pi
   }{3}\right)^2+\left(\frac{8}{3}+\frac{1}{\sqrt{3}}\right) \left(x-\frac{\pi
   }{3}\right)^3+\left(\frac{1}{12}+\frac{4}{\sqrt{3}}\right) \left(x-\frac{\pi
   }{3}\right)^4+O\left(\left(x-\frac{\pi }{3}\right)^5\right)$$ Using again series reversion
$$x=\frac{\pi }{3}+t+\left(5-\frac{7 \sqrt{3}}{2}\right) t^2+\left(\frac{377}{3}-71
   \sqrt{3}\right) t^3+\left(3486-\frac{8069 \sqrt{3}}{4}\right)
   t^4+O\left(t^5\right)$$ where $t=\frac{1}{2} \left(\sqrt{3}-2\right) \left(f(x)-2 \sqrt{3}+3\right)$. This gives as an estimate
$$x=\frac{\pi }{3}+\frac{9}{64} \left(523823248-302429493 \sqrt{3}\right)=1.10580$$
while the solution given by Newton method is $x=1.10582$.
