General Solution Problem

[Scremin34Egl]

2-sinx*cosx-3cos²x = 0

3cos²x + sinxcosx - 2 = 0 ...........What to do from here?

 

[DrPhil]

2-sinx*cosx-3cos²x = 0

3cos²x + sinxcosx - 2 = 0 ...........What to do from here?

Have you learned double-angle formulas? I would be tempted to try them, so that the cos^2(x) would become a linear function of cos(2x), and the (sinx cosx) would become a linear function of sin(2x). The advantage would be having only first powers and no cross-terms.

 

[Bob Brown MSEE]

2-sinx*cosx-3cos²x = 0

3cos²x + sinxcosx - 2 = 0 ...........What to do from here?

One set of solutions are cosx = 0, do you see why?

EDIT: Opps DISREGUARD, got a message that I missed the term 2. Thx Dr. Phil!

 

[Scremin34Egl]

Have you learned double-angle formulas? I would be tempted to try them, so that the cos^2(x) would become a linear function of cos(2x), and the (sinx cosx) would become a linear function of sin(2x). The advantage would be having only first powers and no cross-terms.

Yes, I have, but how do I convert cos²x and sinxcosx to double angle?, maybe multiply by 2?

 

[DrPhil]

Yes, I have, but how do I convert cos²x and sinxcosx to double angle?, maybe multiply by 2?

\(\displaystyle \cos{2A} = 2\cos^2{A} - 1\)

\(\displaystyle \sin{2A} = 2\sin{A}\cos{A}\)

 

[Scremin34Egl]

\(\displaystyle \cos{2A} = 2\cos^2{A} - 1\)

\(\displaystyle \sin{2A} = 2\sin{A}\cos{A}\)

If I multiply the entire equation by 2, I get, 6cos^2*x + 2sinxcosx -4 = 0, I see where the sin2x is coming from but not the 2cos^2x-1. Maybe I shouldn't multiply it by 2?

 

[DrPhil]

If I multiply the entire equation by 2, I get, 6cos^2*x + 2sinxcosx -4 = 0, I see where the sin2x is coming from but not the 2cos^2x-1. Maybe I shouldn't multiply it by 2?

\(\displaystyle \cos{2A} = 2\cos^2{A} - 1 \implies \cos^2A = \frac{1}{2}(\cos{2A} + 1)\)

\(\displaystyle \sin{2A} = 2 \sin{A}\cos{A} \implies \sin{A}\cos{A} = \frac{1}{2} \sin{2A}\)