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.
 A: $$\sin\alpha\sin3\theta+\cos\alpha\cos3\theta-m\cos^3\theta=0$$
$$\sin\alpha\cos3\theta-\cos\alpha\sin3\theta-m\sin^3\theta=0$$
$$-\dfrac{\sin\alpha}{m(\cos^3\theta\sin3\theta+\cos3\theta\sin^3\theta)}=-\dfrac{\cos\alpha}{m(\cos3\theta\cos^3\theta-\sin3\theta\sin^3\theta)}=-1$$
Using $\sin3\theta,\cos3\theta$ formula
$\dfrac{4\sin\alpha}m=(\cos3\theta+3\cos\theta)\sin3\theta+\cos3\theta(3\sin\theta-\sin3\theta)=3\sin(3\theta+\theta)$
Similarly, $\dfrac{4\cos\alpha}m=(\cos3\theta+3\cos\theta)\cos3\theta-\sin3\theta(3\sin\theta-\sin3\theta)=1-3\cos(3\theta+\theta)$
Use $\cos^24\theta+\sin^24\theta=1$
A: Hint for your last equation:
$$\sin(2\theta)=2sin(\theta)\cos(\theta)$$
so $$\sin(2\theta)^2=4\sin^2(\theta)\cos^2(\theta)$$
and you can eliminate $\theta$
So you get $$4\left(1-\cos(\theta)^2\right)\cos(\theta)^2=\frac{4}{3}\frac{m^2-1}{m}$$
Solve this for $\cos(\theta)$
I know this, when you get $\theta$ then you can compute $\alpha$ with the equations above!
Expanding the term
$$\cos(\alpha-3\theta)\sin(\theta)^3-\sin(\alpha-3\theta)\cos(\theta)^3$$
we get
$$4\, \left( \sin \left( \theta \right)  \right) ^{3}\cos \left( \alpha
 \right)  \left( \cos \left( \theta \right)  \right) ^{3}-3\, \left( 
\sin \left( \theta \right)  \right) ^{3}\cos \left( \alpha \right) 
\cos \left( \theta \right) +4\, \left( \sin \left( \theta \right) 
 \right) ^{4}\sin \left( \alpha \right)  \left( \cos \left( \theta
 \right)  \right) ^{2}- \left( \sin \left( \theta \right)  \right) ^{4
}\sin \left( \alpha \right) -4\, \left( \cos \left( \theta \right) 
 \right) ^{6}\sin \left( \alpha \right) +3\, \left( \cos \left( \theta
 \right)  \right) ^{4}\sin \left( \alpha \right) +4\, \left( \cos
 \left( \theta \right)  \right) ^{5}\cos \left( \alpha \right) \sin
 \left( \theta \right) - \left( \cos \left( \theta \right)  \right) ^{
3}\cos \left( \alpha \right) \sin \left( \theta \right)=0 
$$
This term can we solve for $$\alpha$$
$$\alpha=-\arctan \left( 3\,{\frac {\cos \left( \theta \right) \sin \left( 
\theta \right)  \left( 2\, \left( \sin \left( \theta \right)  \right) 
^{2}-1 \right) }{6\, \left( \sin \left( \theta \right)  \right) ^{4}-6
\, \left( \sin \left( \theta \right)  \right) ^{2}+1}} \right) 
$$ and now you can use your $$\theta$$
A: 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.
