I am stuck trying to solve for $A$ in
$3 = 11\sin^2 A - 2\sin 2A$
I cannot see a way to manipulate to get like terms and hence factor it.
Thanks!
I am stuck trying to solve for $A$ in
$3 = 11\sin^2 A - 2\sin 2A$
I cannot see a way to manipulate to get like terms and hence factor it.
Thanks!
$ 11\sin^2 A - 2\sin 2A -3=0$
$ 11\sin^2 A - 4 \sin A \cos A -3\sin^2A -3 \cos^2 A=0$
$ 8\sin^2 A - 4 \sin A \cos A -3 \cos^2 A=0$
Obviously, $\cos(A) \neq 0$ (otherwise $\sin(A)=0$).
Divide by $\cos^2A$ to obtain a quadratic equation in $\tan A$.
You should use the following trigonometric formula
$\sin^2A=\frac{1-\cos(2A)}{2}$
and your equation will become
$\frac{11}{2}-\frac{11}{2}\cos(2A)-2\sin(2A)=3.$
Your next step will be to use the equations
$\cos(2A)=\frac{1-\tan^2A}{1+\tan^2A}$
and
$\sin(2A)=\frac{2\tan A}{1+\tan^2A}$
and will get a quadratic equation for the tangent. This will solve your problem.