Eliminate θ

no one trying........
then let me post the solution

cos(\alpha -3\theta ) = mcos^{3}\theta
cos\alpha cos3\theta +sin\alpha sin3\theta =\frac{m}{4} (cos3\theta +3cos\theta )...........................(1)
sin(\alpha -3\theta )=msin^{3}\theta
sin\alpha cos3\theta -cos\alpha sin3\theta =\frac{m}{4} (3sin\theta -sin3\theta )..........................(2)
from (1) & (2) we get
\Rightarrow sin\alpha = \frac{m}{4}sin3\theta (cos3\theta +3cos\theta )+\frac{m}{4}cos3\theta (3sin\theta -sin3\theta )
\Rightarrow sin\alpha = \frac{3m}{4}sin4\theta \Rightarrow sin4\theta =\frac{4sin\alpha }{3m} ....................................(3)
cos\alpha = \frac{m}{4} cos3\theta (cos3\theta +3cos\theta ) - \frac{m}{4}sin3\theta (3sin\theta -sin3\theta )
\Rightarrow cos\alpha = \frac{m}{4}+\frac{3m}{4}cos4\theta \Rightarrow cos4\theta =\frac{4}{2m}\left(cos\alpha -\frac{m}{4} \right)........(4)

adding (3) and (4) we get
\Rightarrow 1=\frac{16}{9m^{2}}\left\{sin^{2}\alpha +\left(cos\alpha -\frac{m}{4} \right) ^{2}\right\}
\Rightarrow \frac{16}{9m^{2}}+\frac{1}{9}-\frac{8}{9m}cos\alpha = 1
\Rightarrow m^{2}-8mcos\alpha +16=9m^{2}
\Rightarrow m^{2}+mcos\alpha -2=0

HENCE θ is eliminated [1]

ANY MORE METHODS ???????

MORE METHODS WELCOMED [1]