Factoring quadratic equation (with trig identities used)

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SUMMARY

The discussion centers on factoring the quadratic equation of the form \( a \sin^2 x - b \sin 2x + c = 0 \). Participants clarify that the term \( \sin 2x \) can be expressed as \( 2 \sin x \cos x \), complicating the factoring process. A suggested approach involves using trigonometric identities, specifically \( \cos^2 x + \sin^2 x = 1 \) and \( \cos^2 x - \sin^2 x = \cos 2x \), to transform the equation into a solvable format. Ultimately, the equation can be manipulated into a quadratic form in terms of \( \cos 2x \).

PREREQUISITES
  • Understanding of trigonometric identities, specifically \( \sin 2x \) and \( \cos 2x \)
  • Knowledge of quadratic equations and their factoring techniques
  • Familiarity with algebraic manipulation of trigonometric functions
  • Basic skills in solving equations involving trigonometric functions
NEXT STEPS
  • Learn how to apply trigonometric identities in algebraic equations
  • Study the process of transforming trigonometric equations into quadratic forms
  • Explore methods for solving quadratic equations in trigonometric contexts
  • Investigate the implications of using \( \sin^2 x \) and \( \cos^2 x \) in equation solving
USEFUL FOR

Mathematicians, physics students, and educators looking to deepen their understanding of trigonometric equations and their applications in solving quadratic forms.

stratz
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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.
 
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stratz said:
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 =.
 
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).
 
stratz said:
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 <br /> \cos^2 x + \sin ^2 x = 1 \\<br /> \cos^2 x - \sin^2 x = \cos 2x<br /> 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.
 

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