If $\frac{\cos(\alpha -3\theta)}{\cos^3 \theta}=\frac{\sin(\alpha -3\theta)}{\sin^3 \theta}=m$ prove that $\cos\alpha=\frac{2-m^2}{m}$ 
If $$\frac{\cos(\alpha -3\theta)}{\cos^3 \theta}=\frac{\sin(\alpha -3\theta)}{\sin^3 \theta}=m$$
  prove that
  $$\cos\alpha=\frac{2-m^2}{m}$$  

My approach:
$$\cos^2(\alpha-3\theta)+\sin^2(\alpha-3\theta)=m^2(\sin^6\theta+\cos^6\theta)$$
$$\Rightarrow \frac{1}{m^2}=\sin^6\theta+cos^6\theta=1-\frac{3}{4}\sin^22\theta$$
$$\Rightarrow \sin^22\theta=\frac{4}{3}᛫\frac{m^2-1}{m}$$
I can not proceed further, please help.

An approach using factoring expressions is the following: define
 $ \quad x := e^{i\alpha}, \quad y := e^{i\theta}, \quad $ and 
$$ a := \frac{\cos(\alpha -3\theta)}{\cos^3 \theta},
 \, b := \frac{\sin(\alpha -3\theta)}{\sin^3 \theta},
 \, c := \cos \alpha - \frac{2-a^2}a,
 \, d := \cos \alpha - \frac{2-b^2}b, \, e := a-b, $$
$$ F_ 1 \!:=\! 3 x^2 - y^4 + x^2 y^4 - 3 y^8,
 \, F_ 2 \!:=\! 3 x^2 - y^2 + x^2 y^2 - 3 y^4,
 \, F_ 3 \!:=\! 3 x^2 + y^2 - x^2 y^2 - 3 y^4. $$
Now $\, e = 8 F_ 1 y^2/(x (1-y^4)^3) \,$ and thus if $\, F_ 1=0 \,$
then  $\,a=b=m.\,$ Factorization gives
$$ c = F_1F_2/(2 x (1+y^2)^3(x^2+y^6) \quad \text{and}
 \quad d = F_1F_3/(2 x (1-y^2)^3(x^2-y^6))$$ so we have $\, c=d=0\,$
and we are done.
