Solución problema 18

Enunciado del problema 18

Hacemos $se{n}^{2}x=t$. Entonces

$$\frac{1}{t}-\frac{1}{1-t}-\frac{1}{\frac{t}{1-t}}-\frac{1}{\frac{1}{t}}-\frac{1}{\frac{1}{1-t}}-\frac{1}{\frac{t}{1-t}}=$$$$=\frac{1}{t}-\frac{1}{1-t}-\frac{1-t}{t}-t-\left(1-t\right)-\frac{1-t}{t}=$$

$$=\frac{1-t-t-(1-t{)}^2-t(1-t)-(1-t{)}^2}{t\left(1-t\right)}=$$$$=\frac{1-t-t-1+2t-t^2-t+t^2-1+2t-t^2}{t\left(1-t\right)}$$

$$\frac{-t^2+t-1}{t\left(1-t\right)}=-3\Rightarrow -t^2+t-1=-3t+3t^2\Rightarrow 4t^2-4t+1=0$$

$$t=\frac{4\pm \sqrt{16-16}}{8}=\frac{1}{2}\Rightarrow se{n}^{2}x=\frac{1}{2}\Rightarrow \{\begin{array}{cc}x=30º+360k& \\ x=150º+360k& \end{array}$$

Para $k=5,x=1830,x=2190$

Para $k=6,x=1950,x=2310$

Por tanto las soluciones son $2190º$y$1950º$