# Solving trig equation with $\sin A$ and $\sin 2A$

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!

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Hint, use: $\sin^2 A = (1-\cos 2A)/2$. –  Thomas Andrews Dec 16 '11 at 14:55

$$11\sin^2 A - 2\sin 2A -3=0$$

$$11\sin^2 A - 4 \sin A \cos A -3\sin^2A -3 \cos^2 A=0$$

$$8\sin^2 A - 4 \sin A \cos A -3 \cos^2 A=0$$

Obviously, $\cos(A) \neq 0$ (otherwise $\sin(A)=0$).

Divide by $\cos^2A$ to obtain a quadratic equation in $\tan A$.

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Many thanks for an extremely quick answer. Very much appreciated. –  Neil Moffatt Dec 16 '11 at 15:09

You should use the following trigonometric formula

$$\sin^2A=\frac{1-\cos(2A)}{2}$$

$$\frac{11}{2}-\frac{11}{2}\cos(2A)-2\sin(2A)=3.$$

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.

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Many thanks - an elegant solution. –  Neil Moffatt Dec 16 '11 at 15:11