For $0^\circ<x<90^\circ$, the points $P=(\cos x,\cos^2x)$, $Q=(\cos2x,\cos^22x)$, $R=(\cos3x,\cos^23x)$ and $S=(\cos4x,\cos^24x)$ are the vertices of a trapezoid. What is $\cos x$?

1. The four points lie on the parabola $y=u^2$. The chord joining $(a,a^2)$ and $(b,b^2)$ has slope $\dfrac{a^2-b^2}{a-b}=a+b$, so two such chords are parallel exactly when the sums of their $u$-coordinates are equal.
2. Let $c_k=\cos kx$. The trapezoid has one pair of opposite sides parallel; the three possibilities are
   \begin{gather*}PQ\parallel RS:\ c_1+c_2=c_3+c_4,\\ PR\parallel QS:\ c_1+c_3=c_2+c_4,\\ PS\parallel QR:\ c_1+c_4=c_2+c_3.\end{gather*}
3. By the sum-to-product formula, $c_1+c_2=2\cos\frac{3x}2\cos\frac x2$ and $c_3+c_4=2\cos\frac{7x}2\cos\frac x2$, so the first case needs $\cos\frac{3x}2=\cos\frac{7x}2$, i.e. $x=72^\circ$ in $(0^\circ,90^\circ)$. The second case gives $2\cos x\,(\cos2x-\cos3x)=0$, again $x=72^\circ$. But at $x=72^\circ$, $\cos144^\circ=\cos216^\circ$, so $Q=R$ and there is no trapezoid.
4. So $PS\parallel QR$: $c_1+c_4=2\cos\frac{5x}2\cos\frac{3x}2$ and $c_2+c_3=2\cos\frac{5x}2\cos\frac x2$, hence $\cos\frac{5x}2\left(\cos\frac{3x}2-\cos\frac x2\right)=0$. The second factor vanishes only at $x=0$, so $\cos\frac{5x}2=0$ and $x=36^\circ$ (the next value $108^\circ$ is too large).
5. Finally $\cos36^\circ=\boxed{\dfrac{1+\sqrt5}4}$ (with $c=\cos36^\circ$, $\cos108^\circ=-\cos72^\circ$ gives $4c^3-3c=-(2c^2-1)$, so $(c+1)(4c^2-2c-1)=0$ and $c=\frac{1+\sqrt5}4>0$).