Use the angle sum identities to expand [tex]\sin(3A)[/tex] and [tex]\cos(3A)[/tex]:
[tex]\sin(3A)=\sin(A+2A)=\sin A\cos(2A)+\cos A\sin(2A)[/tex]
[tex]\cos(3A)=\cos(A+2A)=\cos A\cos(2A)-\sin A\sin(2A)[/tex]
Use the double angle identity to expand [tex]\sin(2A)[/tex]:
[tex]\sin(2A)=2\sin A\cos A[/tex]
So we have
[tex]\sin(3A)=\sin A\cos(2A)+2\cos^2A\sin A[/tex]
[tex]\cos(3A)=\cos A\cos(2A)-2\sin^2A\cos A[/tex]
Then divide the first expression by [tex]\sin A[/tex] and the second by [tex]\cos A[/tex]:
[tex]\dfrac{\sin(3A)}{\sin A}=\cos(2A)+2\cos^2A[/tex]
[tex]\dfrac{\cos(3A)}{\cos A}=\cos(2A)-2\sin^2A[/tex]
Squaring these gives
[tex]\dfrac{\sin^2(3A)}{\sin^2A}=(\cos(2A)+2\cos^2A)^2=\cos^2(2A)+4\cos(2A)\cos^2A+4\cos^4A[/tex]
[tex]\dfrac{\cos^2(3A)}{\cos^2A}=(\cos(2A)-2\sin^2A)^2=\cos^2(2A)-4\cos(2A)\sin^2A+4\sin^4A[/tex]
Subtract the second expression from the first to get the original equation. The [tex]\cos^2(2A)[/tex] terms cancel, leaving us with
[tex]4\cos(2A)\cos^2A+4\cos^4A+4\cos(2A)\sin^2A-4\sin^4A=2[/tex]
Now, notice that
[tex]4\cos(2A)\cos^2A+4\cos(2A)\sin^2A=4\cos(2A)(\cos^2A+\sin^2A)=4\cos(2A)[/tex]
and
[tex]4\cos^4A-4\sin^4A=4(\cos^2A-\sin^2A)(\cos^2A+\sin^2A)=4\cos(2A)[/tex]
because [tex]\cos^2A+\sin^2A=1[/tex] and [tex]\cos^2A-\sin^2A=\cos(2A)[/tex].
So we're left with
[tex]4\cos(2A)+4\cos(2A)=8\cos(2A)=2\implies\boxed{\cos(2A)=\dfrac14}[/tex]
