Resolve: $4\sin(2x)+4\cos(x)-5=0$

Question

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.

Answer 1

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.

Answer 2

I want to release the user's hint.

Let $\tan\frac{x}{2}=t$.

Thus, $$\sin\frac{x}{2}=\frac{2t}{1+t^2},$$ $$\cos\frac{x}{2}=\frac{1-t^2}{1+t^2}$$

and we need to solve that $$\frac{16t(1-t^2)}{(1+t^2)^2}+\frac{4(1-t^2)}{1+t^2}-5=0$$ or $$9t^4+16t^3+10t^2-16t+1=0$$ or for any $\alpha$ $$(1-8t+\alpha t^2)^2-t^2((\alpha^2-9)t^2-16(\alpha+1)t+2a+54)=0.$$ Now, we'll choose a value of $\alpha$ such that $\alpha^2-9>0$, $\alpha+27>0$ and $$64(\alpha+1)^2-(2\alpha+54)(\alpha^2-9)=0$$ or $$\alpha^3-5\alpha^2-73\alpha-275=0,$$ which gives $$\alpha=\frac{1}{3}\left(5+2\sqrt[3]{685-6\sqrt{6729}}+2\sqrt[3]{685+6\sqrt{6729}}\right)$$ and for this value of $\alpha$ we need to solve $$(1-8t+\alpha t^2)^2-t^2\left(\sqrt{\alpha^2-9}t-\frac{8(\alpha+1)}{\sqrt{\alpha^2-9}}\right)^2=0$$ or $$\left(\left(\sqrt{\alpha^2-9}-\alpha\right)t^2-8\left(\tfrac{\alpha+1}{\sqrt{\alpha^2-9}}-1\right)t-1\right)\left(\left(\sqrt{\alpha^2-9}+\alpha\right)t^2-8\left(\tfrac{\alpha+1}{\sqrt{\alpha^2-9}}+1\right)t+ 1\right)=0.$$ Easy to see that the equation $$\left(\sqrt{\alpha^2-9}-\alpha\right)t^2-8\left(\tfrac{\alpha+1}{\sqrt{\alpha^2-9}}-1\right)t- 1=0$$ has no real roots, but the second equation $$\left(\sqrt{\alpha^2-9}+\alpha\right)t^2-8\left(\tfrac{\alpha+1}{\sqrt{\alpha^2-9}}+1\right)t+1=0$$ gives $$t=\frac{4\left(\frac{\alpha+1}{\sqrt{\alpha^2-9}}+1\right)+\sqrt{16\left(\frac{\alpha+1}{\sqrt{\alpha^2-9}}+1\right)^2-\sqrt{\alpha^2-9}-\alpha}}{\sqrt{\alpha^2-9}+\alpha}$$ or $$t=\frac{4\left(\frac{\alpha+1}{\sqrt{\alpha^2-9}}+1\right)-\sqrt{16\left(\frac{\alpha+1}{\sqrt{\alpha^2-9}}+1\right)^2-\sqrt{\alpha^2-9}-\alpha}}{\sqrt{\alpha^2-9}+\alpha},$$ which gives the answer: $$x=2\arctan\frac{4\left(\frac{\alpha+1}{\sqrt{\alpha^2-9}}+1\right)+\sqrt{16\left(\frac{\alpha+1}{\sqrt{\alpha^2-9}}+1\right)^2-\sqrt{\alpha^2-9}-\alpha}}{\sqrt{\alpha^2-9}-\alpha}+2\pi k$$ or

$$x=2\arctan\frac{4\left(\frac{\alpha+1}{\sqrt{\alpha^2-9}}+1\right)-\sqrt{16\left(\frac{\alpha+1}{\sqrt{\alpha^2-9}}+1\right)^2-\sqrt{\alpha^2-9}-\alpha}}{\sqrt{\alpha^2-9}+\alpha}+2\pi k,$$ where $k$ is an integer number and $\alpha=\frac{1}{3}\left(5+2\sqrt[3]{685-6\sqrt{6729}}+2\sqrt[3]{685+6\sqrt{6729}}\right).$

Answer 3

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 $x=1.10582$.