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.