Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

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

2 Answers 2

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

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

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

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

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

share|improve this answer
    
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

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

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.

share|improve this answer
    
Many thanks - an elegant solution. –  Neil Moffatt Dec 16 '11 at 15:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.