Ytukyg

Let $alpha=frac{180^o}7$.

Consider that

$$sin(3alpha)=sin(180^o-3alpha)=sin(4alpha)$$

and

$$cos(3alpha)=-cos(180^o-3alpha)=-cos(4alpha)text{.}$$

From $sin(3alpha)=sin(4alpha)$, we get

$$sin(alpha)cos(2alpha)+sin(2alpha)cos(alpha)=sin(2alpha)cos(2alpha)+sin(2alpha)cos(2alpha)$$
$$cos(2alpha)(sin(alpha)-sin(2alpha))=sin(2alpha)(cos(2alpha)-cos(alpha))$$
$$cos(2alpha)sin(alpha)(1-2cos(alpha))=2sin(alpha)cos(alpha)(2cos^2(alpha)-cos(alpha)-1)$$
$$cos(2alpha)(1-2cos(alpha))=2cos(alpha)(2cos^2(alpha)-cos(alpha)-1)$$
$$cos(2alpha)(1-2cos(alpha))=-2cos^2(alpha)(1-2cos(alpha))-2cos(alpha)$$
$$(cos(2alpha)+2cos^2(alpha))(1-2cos(alpha))=-2cos(alpha)$$
$$(cos(2alpha)+2cos^2(alpha))(2cos(alpha)-1)=2cos(alpha)$$
$$(4cos^2(alpha)-1)(2cos(alpha)-1)=2cos(alpha)$$
$$8cos^3(alpha)-4cos^2(alpha)-4cos(alpha)+1=0$$

From $cos(3alpha)=-cos(4alpha)$ we get

$$cos(2alpha)cos(alpha)-sin(2alpha)sin(alpha)=sin(2alpha)sin(2alpha)-cos(2alpha)cos(2alpha)$$
$$cos(2alpha)(cos(2alpha)+cos(alpha))=sin(2alpha)(sin(2alpha)+sin(alpha))$$
$$(2cos^2(alpha)-1)(2cos^2(alpha)+cos(alpha)-1)=sin^2(2alpha)+sin(2alpha)sin(alpha)$$
$$(2cos^2(alpha)-1)(2cos^2(alpha)+cos(alpha)-1)=1-cos^2(2alpha)+2sin^2(alpha)cos(alpha)$$
$$(2cos^2(alpha)-1)(2cos^2(alpha)+cos(alpha)-1)=1-(2cos(alpha)-1)^2+2(1-cos^2(alpha))cos(alpha)$$
$$4cos^4(alpha)+2cos^3(alpha)-4cos^2(alpha)-cos(alpha)+1=1-(4cos^2(alpha)-4cos(alpha)+1)+2cos(alpha)-2cos^3(alpha)$$
$$4cos^4(alpha)+2cos^3(alpha)-4cos^2(alpha)-cos(alpha)=-4cos^2(alpha)+4cos(alpha)-1+2cos(alpha)-2cos^3(alpha)$$
$$4cos^4(alpha)+4cos^3(alpha)-7cos(alpha)+1=0$$

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):

$$4c^2-12c+7=0$$
$$4c^2-12c+9=2$$
$$(2c-3)^2=2$$
$$2c-3=pmsqrt2$$
$$2c=3pmsqrt2$$
$$c=frac{3pmsqrt2}2$$
$$cos(frac{180^o}7)=frac{3pmsqrt2}2$$

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 $cos(3alpha)=-cos(4alpha)$, I made the mistake of turning $cos(2alpha)$ to $2cos(alpha)-1$, instead of the correct $2cos^2(alpha)-1$.

Now let's carry on, fixing the mistake:

$$(2cos^2(alpha)-1)(2cos^2(alpha)+cos(alpha)-1)=1-(2cos^2(alpha)-1)^2+2(1-cos^2(alpha))cos(alpha)$$
$$4cos^4(alpha)+2cos^3(alpha)-4cos^2(alpha)-cos(alpha)+1=1-(4cos^4(alpha)-4cos^2(alpha)+1)+2cos(alpha)-2cos^3(alpha)$$
$$4cos^4(alpha)+2cos^3(alpha)-4cos^2(alpha)-cos(alpha)=-4cos^4(alpha)+4cos^2(alpha)-1+2cos(alpha)-2cos^3(alpha)$$
$$8cos^4(alpha)+4cos^3(alpha)-8cos^2(alpha)-3cos(alpha)+1=0$$

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).