If $\alpha$, $\beta$, $\gamma$, the angles made by a vector with $x, y, z$ axes, are in the ratio 1:2:3 then their values are? This is a question asked in a book:

If $\alpha$, $\beta$, $\gamma$ are angles made by a vector $x, y, z$ axes respectively, and $\alpha$:$\beta$:$\gamma$ = 1:2:3 then values of $\alpha$, $\beta$, $\gamma$ are?

I tried the obvious by taking  $\alpha$ = k; $\beta$=2k; $\gamma$ = 3k; and substituting in  $cos^2\alpha$+$cos^2\beta$+$cos^2\gamma$ = 1
but its complicating things. I'm must be missing something simple and obvious but I can't figure out what. 

If
$\cos^2 x +\cos^2(2x)+\cos^2(3x)=1
$,
then,
since
$\cos^2(z)
=(\cos(2z)+1)/2
$,
$1
=(\cos(2x)+\cos(4x)+\cos(6x)+3)/2
$
or
$\cos(2x)+\cos(4x)+\cos(6x)
= -1
$.
Letting
$y = 2x$,
$\begin{array}\\
-1
&=\cos(y)+\cos(2y)+\cos(3y)\\
&=\cos(y)+2\cos^2y-1+4\cos^3(y)-3\cos(y)\\
&=4\cos^3(y)+2\cos^2y-2\cos(y)-1\\
\end{array}
$
or
$4\cos^3(y)+2\cos^2y-2\cos(y)
=0
$.
Letting
$z = \cos y$,
$\begin{array}\\
0
&=2z^3+2z^2-z\\
&=z(2z^2+z-1)\\
&=z(z+1)(2z-1)
\qquad\text{just fixed an error here}\\
\end{array}
$
so
$z = 0, -1, \frac12
$.
Therefore
$y=\pi/2, \pi, \pi/3$
so
$x
= y/2
=\pi/4, \pi/2, \pi/6
$.
