Asked by Stevium
Tan theta +2sec theta=2 cot theta+3cosec theta=a Find theta
Answers
Answered by
Steve
I'll just use x, for ease of typing
tanx + 2secx = 2cotx + 3cscx
A strange mixture of functions, so I'll just use sine and cos:
sinx/cosx + 2/cosx = 2cosx/sinx + 3/sinx
(sinx+2)/cosx = (2cosx+3)/sinx
So, assuming that cosx and sinx are nonzero,
sinx(sinx+2)-cosx(2cosx+3) = 0
sin^2x+2sinx - 2cos^2x-3cosx = 0
1-cos^2x + 2sinx - 2cos^2x - 3cosx = 0
3cos^2x +3cosx - 1 = 2sinx
That's not gonna be an easy one. I doubt that it will factor, so a graphic method will be your best bet.
tanx + 2secx = 2cotx + 3cscx
A strange mixture of functions, so I'll just use sine and cos:
sinx/cosx + 2/cosx = 2cosx/sinx + 3/sinx
(sinx+2)/cosx = (2cosx+3)/sinx
So, assuming that cosx and sinx are nonzero,
sinx(sinx+2)-cosx(2cosx+3) = 0
sin^2x+2sinx - 2cos^2x-3cosx = 0
1-cos^2x + 2sinx - 2cos^2x - 3cosx = 0
3cos^2x +3cosx - 1 = 2sinx
That's not gonna be an easy one. I doubt that it will factor, so a graphic method will be your best bet.
Answered by
Reiny
tanØ + 2secØ = 2cotØ + 3cscØ
sinØ/cosØ + 2/cosØ = 2cosØ/sinØ + 3/sinØ
multiply each term by sinØcosØ
sin^2 Ø + 2sinØ = 2cos^2 Ø + 3cosØ
sinØ(sinØ + 2) = cosØ(2cosØ + 3)
new start: I will use x instead of Ø for easier typing
2secx - 3cscx = 2cotx - tanx
2/cosx - 3/sinx = 2cosx/sinx - sinx/cosx
(2sinxcosx - 3cosx)/(sinxcosx = (2cos^2 x - sin^2 x)/(sinxcosx)
2sinxcosx - 3cosx = 2cos^2 x - sin^2 x
cosx(2sinx - 3) = 2(1 - sin^2 x) - sin^2 x
cosx(2sinx - 3) = 2 - 3sin^2 x
While Steve was posting his solution, I must have been working on mine.
I reached the same stage as he did, but then tried another approach, again getting real messy
I tried Wolfram, and got this mess.
http://www.wolframalpha.com/input/?i=solve+y+%3D+tanx%2B+2secx++%3D+2cotx+%2B+3cscx
sinØ/cosØ + 2/cosØ = 2cosØ/sinØ + 3/sinØ
multiply each term by sinØcosØ
sin^2 Ø + 2sinØ = 2cos^2 Ø + 3cosØ
sinØ(sinØ + 2) = cosØ(2cosØ + 3)
new start: I will use x instead of Ø for easier typing
2secx - 3cscx = 2cotx - tanx
2/cosx - 3/sinx = 2cosx/sinx - sinx/cosx
(2sinxcosx - 3cosx)/(sinxcosx = (2cos^2 x - sin^2 x)/(sinxcosx)
2sinxcosx - 3cosx = 2cos^2 x - sin^2 x
cosx(2sinx - 3) = 2(1 - sin^2 x) - sin^2 x
cosx(2sinx - 3) = 2 - 3sin^2 x
While Steve was posting his solution, I must have been working on mine.
I reached the same stage as he did, but then tried another approach, again getting real messy
I tried Wolfram, and got this mess.
http://www.wolframalpha.com/input/?i=solve+y+%3D+tanx%2B+2secx++%3D+2cotx+%2B+3cscx
Answered by
Bane
You guys are interpreting the question wrong, there are two equations in the question and your to find the value of a the first equ ending in 2 and the 2nd starting from coy
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