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Solve the following equation:

Posted January 13th, 2010 by alpha
in

I've trouble with the this question.
$ sin^3\theta cos\theta-cos^3\theta sin\theta=\frac{1}{4} $

‹ Find the general solution PROVE THAT ›

Comments

I've just solved this way-

On January 13th, 2010 alpha says:

But my answer is not matching with the answer written in the book.
$ 4sin^3\theta cos\theta-4cos^3\theta sin\theta -1=0 $
$ 4sin\theta cos\theta(sin^2\theta -cos^2 \theta)-1=0 $
$ 4sin\theta cos\theta (-cos2\theta)-1=0 $
$ -2 sin2\theta cos2\theta-1=0 $
$ sin4\theta= -1 $
$ 4\theta=n \pi \pm(-1)^n (-\frac{\pi}{2}) $
-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

answer

On January 13th, 2010 Structure says:

$ 4\theta=n \pi \pm(-1)^n (-\frac{\pi}{2}) $
There is no $ \pm $ in front of $ (-1)^n $
Solution is $ 4\theta=n \pi +(-1)^n (-\frac{\pi}{2}) $
This solution is the same with
$ 4\theta=2n \pi -\frac{\pi}{2}=2(n-1)\pi+\frac{3\pi}{2}\; n\in Z $

Unfortunately the answer given in the book is different.

On January 13th, 2010 alpha says:

It is
$ \frac{n\pi}{2}+\frac{3\pi}{8} $

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

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