i hope that DEACART'S RULE OF SIGN will be helpful to solve it .

please try it . and then tell me what is the result .

After full expanding we get $ 128x^7-192x^5+80x^3-8x-1=0 $ so if there's a rational solution by the rational root theorem we have that they are in the form $\frac{a}{b}$ where $a$ is a divisor if $\pm1$ and $b$ is a divisor of $\pm{128} $ but $128$ is $2^7$ so the solutions can be in the form $ \pm \frac{1}{2^n} $ but since we have to find in the interval $[0,1]$ we know they are positive so we only find in the form $x=\frac{1}{2^n}$. So plugging $x=\frac{1}{2^n}$ we get $ 2^{7-7n}-2^{6-5n}*3+2^{4-3n}*5-2^{3-n}=1 $ but since $LHS$ must be equal to $1$ we need one of the $ 2^i$ to be equal to $1$ (so $i=0$) becouse if not the equation has no solutions by checking mod2, so we have some cases: $7-7n=0$ or $6-5n=0$ or $4-3n=0$ or $3-n=0$, but we have $n\in N$ so $6-5n=0$ and $4-3n=0$ no solutions, still to check $7-7n=0$ and $3-n=0$ for where we find the solutions $n=1$ or $n=3$ yelds $x=\frac{1}{2}$ or $x=\frac{1}{8} $, plug in original equation to get $x=\frac{1}{2}$ solution. I can't find a way to find other solutions in that interval

yes good solution .

hint

CantonMathGuy wrote:

hint

True approach

CantonMathGuy wrote:

hint

Really nice dear thank you! I'll only end with ur hint. Set $ x=cos(y) $ the equation becomes $ 8cos(y)(2cos(y)^2-1)(8cos(y)^4-8cos(y)^2+1)=1 $ but we knwo $ 2cos(y)^2-1=cos(2y) $ and $ 8cos(y)^4-8cos(y)^2+1=cos(4y) $ so the equation becomes $ cos(y)cos(2y)cos(4y)=\frac{1}{8} $. Now let $ w=cos(y)+isin(y) $ we have that $ cos(y)=\frac{w+\frac{1}{w}}{2} $ and $ cos(2y)=\frac{w^2+\frac{1}{w^2}}{2} $ and $ cos(4y)=\frac{w^4+\frac{1}{w^4}}{2} $ plugging this in our equation it becomes: $ (w+\frac{1}{w})(w^2+\frac{1}{w^2})(w^4+\frac{1}{w^4})=1 $ now clearing denominators we have $ w^{14}+w^{12}+w^{10}+w^8-w^7+w^6+w^4+w^2+1=0 $ but this is equivalent to $ (w^2-w+1)(w^6-w^3+1)(w^6+w^5+w^4+w^3+w^2+w+1)=0 $, now a little thing before solving: we have to find the solution in the interval $[0,1]$ and we know that if $ w_0 $ is a solution for our euqation then it's also $ \overline{w_0} $ but $ w$ and $w_0$ have the same $cos$ so we just find the complex solution of our equation in the first quadrant. 1)From $ w^2-w+1=0 $ it's well know $ w_1=cos(\pi/3)+isin(pi/3) $ 2)From $ w^6-w^3+1=0 $ setting $ z=w^3 $ we have the equation 1) so solutions $ w_2=cos(\pi/9)+isin(\pi/9) $ 3)From $w^6+w^5+w^4+w^3+w^2+w+1=0 $ factor $ \frac{w^7-1}{w-1}=0 $ we need $ w^7-1=0$ with $ w\neq1 $ here we find $w_3=cos(2\pi/7)+isin(2\pi/7) $. Now solutions are just $ x_1=Re(w_1)=\frac{1}{2} $ , $x_2=Re(w_2)=cos(\pi/9) $ and $x_3=Re(w_3)=cos(2\pi/7) $