Determine the range of a function

Question

Find the range of the following function : $$y= (3\cos{2x} -\sin{2x} -1)^2 -8$$ First solution: $$ (3\cos{2x} -\sin{2x} -1)^2= (\sqrt{10} \cos{2x+ \theta})^2$$ $$-\sqrt{10} =<(\sqrt{10} \cos{(2x+ \theta}))=< \sqrt{10}$$ $$-1-\sqrt{10} =<(\sqrt{10} \cos{(2x+ \theta}))-1 =< \sqrt{10} -1$$ $$-8=< \left((\sqrt{10} \cos{(2x+ \theta}))-1\right)^2 -8=< 3-2\sqrt{10}$$ Second solution: $$ (3\cos{2x} -\sin{2x} -1)^2=(2\sin{2x}-3\cos{2x} +1)^2 = (\sqrt{10} \sin\left(2x -\theta\right))^2 $$ Similarly as the first we get $$-8 =< y =< 3+2\sqrt{10}$$ Which solution is correct ? and whats the mistake ?

Answer 1

$$3\cos2x-2\sin2x=\sqrt{3^2+1^2}\cos\left(2x+\arccos\dfrac3{\sqrt{10}}\right)$$

$\implies-\sqrt{10}-1\le3\cos2x-\sin2x-1\le\sqrt{10}-1$

$$3\cos2x-\sin2x-1\le\sqrt{10}-1\implies(3\cos2x-\sin2x-1)^2-8\le(\sqrt{10}-1)^2-8$$

$$3\cos2x-\sin2x-1\ge-\sqrt{10}-1\implies(3\cos2x-\sin2x-1)^2-8\le(-\sqrt{10}-1)^2-8$$

Take the union of the two ranges