Solving Trigonometry Problem: 2 Variables, 2 Equations

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The discussion revolves around solving a two-variable trigonometry problem involving two equations. The first equation is 3 = 1.2(cosx) + v(cos30), and the second is 0 = 1.2(sinx) - v(sin30). Participants suggest solving for v from the second equation and substituting it into the first, but there is confusion regarding the correct manipulation and interpretation of the equations. Some participants propose using identities and relationships between sine and cosine to simplify the problem further. The conversation highlights the challenges in solving simultaneous trigonometric equations and the need for clarity in applying trigonometric identities.
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I've been really stumped by this one from my math class. It's a two variable trig. question with two equations. I believe there simultaneous but I'm not sure

3 = 1.2(cosx) + v(cos30)
0 = 1.2(sinx) - v(sin30)

Solve for the values of x & v

Please explain how you got your answer or show work. Thanks
 
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Well, solve for v first from, say, the 2nd eq.
Then you get a trig equation; use the summation formula to reshape it so that you can solve for relevant x's
 
Sorry I'm not familar with the summation formula. I solved for V in the second equation and got V = .6sinx.

I substituted that into the second equation and got sin = 0, 180 or 360 but then the 2nd equation doesn't check out. I know I'm doing it wrong somehow just not sure how
 
Champ07 said:
I've been really stumped by this one from my math class. It's a two variable trig. question with two equations. I believe there simultaneous but I'm not sure

3 = 1.2(cosx) + v(cos30)
0 = 1.2(sinx) - v(sin30)

Solve for the values of x & v

Please explain how you got your answer or show work. Thanks
Actually, I suspect the problem asked YOU to explain how you got your answer or shiow work. I don't have to! Let x= cos(x) and y= sin(x). then you have the two equations
3= 1.2x+ vy and 0 = 1.2x+ vy as well as the obvious equation x^2+ y^2= 1. What happens if you add the first two equations?
 
HallsofIvy how do you get those equations?

Seems it should be
<br /> 3 = 1.2x + v\frac{\sqrt{3}}{2}
0 = 1.2y - 0.5v
x^2 + y^2 = 1<br />
 
How did I get those two equations? I copied them from the first post in this thread!
3 = 1.2(cosx) + v(cos30)
0 = 1.2(sinx) - v(sin30)
I now see, looking more closely, that it might be good idea to multiply the first equation by sin(30) and the second equation by cos(30). Although it wasn't said, I assume that is "30 degrees" so that cos(30)= \sqrt{2}/3 and sin(30)= 1/2.
 
Champ07 said:
Sorry I'm not familar with the summation formula. I solved for V in the second equation and got V = .6sinx.

That's the right idea, but the solution to 0 = 1.2(sinx) - v(sin30) = 1.2(sinx) -v(0.5) isn't v = 0.6 sin x.
 
i just solved it, long prob
 

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