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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.


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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


and your equation will become


Your next step will be to use the equations



$$\sin(2A)=\frac{2\tan A}{1+\tan^2A}$$

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

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