Factoring quadratic equation (with trig identities used)

[stratz]

Is it possible to factor a quadratic equation along the lines of asin^2x -bsin2x+c ? If so, how? The sin2x seems to be a problem since when expanded it becomes 2sinxcosx, but I'm wondering if it is possible, and how it would be done?

Thanks in advance.

 

[Mark44]

Is it possible to factor a quadratic equation along the lines of asin^2x -bsin2x+c ? If so, how? The sin2x seems to be a problem since when expanded it becomes 2sinxcosx, but I'm wondering if it is possible, and how it would be done?

I don't see any way of doing it. What's the context of this problem? Could there be a mistake leading up to what you show?

BTW, what you show isn't an equation -- there's no =.

 

[mfb]

a sin²(x) - b sin(2x) + c = 0 (I guess that is what you meant) can be written as $a \sin^2(x) + c = 2 b \sin(x) \sqrt{1-\sin^2(x)}$, after squaring both sides you get a quadratic equation in sin²(x).

 

[pasmith]

Is it possible to factor a quadratic equation along the lines of asin^2x -bsin2x+c ? If so, how? The sin2x seems to be a problem since when expanded it becomes 2sinxcosx, but I'm wondering if it is possible, and how it would be done?

Thanks in advance.

Is the object to solve $a \sin^2x - b \sin 2x + c = 0$ for $x$? If so, use the identities $$\cos^2 x + \sin ^2 x = 1 \\<br /> \cos^2 x - \sin^2 x = \cos 2x$$ to express $\sin^2 x$ in terms of $\cos 2x$; then you'll have something of the form $p \cos 2x - b \sin 2x + q = 0$ which I hope you know how to solve.