Let $\alpha=\frac{180^o}7$.
Consider that
$$\sin(3\alpha)=\sin(180^o-3\alpha)=\sin(4\alpha)$$
and
$$\cos(3\alpha)=-\cos(180^o-3\alpha)=-\cos(4\alpha)\text{.}$$
From $\sin(3\alpha)=\sin(4\alpha)$, we get
$$\sin(\alpha)\cos(2\alpha)+\sin(2\alpha)\cos(\alpha)=\sin(2\alpha)\cos(2\alpha)+\sin(2\alpha)\cos(2\alpha)$$ $$\cos(2\alpha)(\sin(\alpha)-\sin(2\alpha))=\sin(2\alpha)(\cos(2\alpha)-\cos(\alpha))$$ $$\cos(2\alpha)\sin(\alpha)(1-2\cos(\alpha))=2\sin(\alpha)\cos(\alpha)(2\cos^2(\alpha)-\cos(\alpha)-1)$$ $$\cos(2\alpha)(1-2\cos(\alpha))=2\cos(\alpha)(2\cos^2(\alpha)-\cos(\alpha)-1)$$ $$\cos(2\alpha)(1-2\cos(\alpha))=-2\cos^2(\alpha)(1-2\cos(\alpha))-2\cos(\alpha)$$ $$(\cos(2\alpha)+2\cos^2(\alpha))(1-2\cos(\alpha))=-2\cos(\alpha)$$ $$(\cos(2\alpha)+2\cos^2(\alpha))(2\cos(\alpha)-1)=2\cos(\alpha)$$ $$(4\cos^2(\alpha)-1)(2\cos(\alpha)-1)=2\cos(\alpha)$$ $$8\cos^3(\alpha)-4\cos^2(\alpha)-4\cos(\alpha)+1=0$$
From $\cos(3\alpha)=-\cos(4\alpha)$ we get
$$\cos(2\alpha)\cos(\alpha)-\sin(2\alpha)\sin(\alpha)=\sin(2\alpha)\sin(2\alpha)-\cos(2\alpha)\cos(2\alpha)$$ $$\cos(2\alpha)(\cos(2\alpha)+\cos(\alpha))=\sin(2\alpha)(\sin(2\alpha)+\sin(\alpha))$$ $$(2\cos^2(\alpha)-1)(2\cos^2(\alpha)+\cos(\alpha)-1)=\sin^2(2\alpha)+\sin(2\alpha)\sin(\alpha)$$ $$(2\cos^2(\alpha)-1)(2\cos^2(\alpha)+\cos(\alpha)-1)=1-\cos^2(2\alpha)+2\sin^2(\alpha)\cos(\alpha)$$ $$(2\cos^2(\alpha)-1)(2\cos^2(\alpha)+\cos(\alpha)-1)=1-(2\cos(\alpha)-1)^2+2(1-\cos^2(\alpha))\cos(\alpha)$$ $$4\cos^4(\alpha)+2\cos^3(\alpha)-4\cos^2(\alpha)-\cos(\alpha)+1=1-(4\cos^2(\alpha)-4\cos(\alpha)+1)+2\cos(\alpha)-2\cos^3(\alpha)$$ $$4\cos^4(\alpha)+2\cos^3(\alpha)-4\cos^2(\alpha)-\cos(\alpha)=-4\cos^2(\alpha)+4\cos(\alpha)-1+2\cos(\alpha)-2\cos^3(\alpha)$$ $$4\cos^4(\alpha)+4\cos^3(\alpha)-7\cos(\alpha)+1=0$$
Now, let $c=\cos(\alpha)$. From the previous equations:
$$8c^3-4c^2-4c+1=0\ ...(1)$$ $$4c^4+4c^3-7c+1=0\ ...(2)$$
Subtract the equations:
$$4c^4-4c^3+4c^2-3c=0$$
Since $c$ cannot be $0$ ($c=0$ doesn't fulfill (1)), we get
$$4c^3-4c^2+4c-3=0$$ $$8c^3-8c^2+8c-6=0\ ...(3)$$
Subtract (3) from (1):
$$4c^2-12c+7=0$$ $$4c^2-12c+9=2$$ $$(2c-3)^2=2$$ $$2c-3=\pm\sqrt2$$ $$2c=3\pm\sqrt2$$ $$c=\frac{3\pm\sqrt2}2$$ $$\cos(\frac{180^o}7)=\frac{3\pm\sqrt2}2$$
Since $\cos(\frac{180^o}7)\le1$, we get
$$\cos(\frac{180^o}7)=\frac{3-\sqrt2}2$$
But I substituted it to (1), and it doesn't fulfill the equation.
What did I do wrong?
In the fifth line of the argument from $\cos(3\alpha)=-\cos(4\alpha)$, I made the mistake of turning $\cos(2\alpha)$ to $2\cos(\alpha)-1$, instead of the correct $2\cos^2(\alpha)-1$.
Now let's carry on, fixing the mistake:
$$(2\cos^2(\alpha)-1)(2\cos^2(\alpha)+\cos(\alpha)-1)=1-(2\cos^2(\alpha)-1)^2+2(1-\cos^2(\alpha))\cos(\alpha)$$ $$4\cos^4(\alpha)+2\cos^3(\alpha)-4\cos^2(\alpha)-\cos(\alpha)+1=1-(4\cos^4(\alpha)-4\cos^2(\alpha)+1)+2\cos(\alpha)-2\cos^3(\alpha)$$ $$4\cos^4(\alpha)+2\cos^3(\alpha)-4\cos^2(\alpha)-\cos(\alpha)=-4\cos^4(\alpha)+4\cos^2(\alpha)-1+2\cos(\alpha)-2\cos^3(\alpha)$$ $$8\cos^4(\alpha)+4\cos^3(\alpha)-8\cos^2(\alpha)-3\cos(\alpha)+1=0$$
Now, let $c=\cos(\alpha)$. From the previous equations:
$$8c^3-4c^2-4c+1=0\ ...(1)$$ $$8c^4+4c^3-8c^2-3c+1=0\ ...(2)$$
Subtract (1) from (2):
$$8c^4-4c^3-4c^2+c=0$$
Since $c$ cannot be $0$ ($c=0$ doesn't fulfill (1)), we get
$$8c^3-4c^2-4c+1=0\ ...(1)$$
I can't go further, so the value to find is the root of that equation which is in the interval $(0,1)$ (only one).