Solve the Equation exactly over the interval [0, 2pi] the problems i do not understand are: 2sin^2x + sin x - 1 = 0 2sinx cosx + cos(x) = 0 2sin^2x - sin x = 0 if anyone could help me i would appreciate it alot. thanks
Last edited by mr fantastic; Nov 29th 2011 at 02:50 PM. Reason: Title.
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Originally Posted by trigonometry424 Solve the Equation exactly over the interval [0, 2pi] the problems i do not understand are: 2sin^2x + sin x - 1 = 0 2sinx cosx + cos(x) = 0 2sin^2x - sin x = 0 2sin^2x + sin x - 1 = 0 if anyone could help me i would appreciate it alot. thanks What have you tried? 1) Use the quadratic formula to get values of sin(x) and thus x. 2) Factor out cos(x) 3) Factor out sin(x) 4) Same as 1 If you're stuck post what you've tried
i know how to do problem 1 and 2 now i think i just need to know how to factor the 3rd one now
Originally Posted by trigonometry424 i know how to do problem 1 and 2 now i think i just need to know how to factor the 3rd one now Unless I'm missing something equation 4 is the same as equation 1 so you've solved that too. Number 3: You can factor out sin(x) since it's a common factor: Either or Can you find x in those cases?
yea sorry it was an accident i edited my op and would x = 1/2?
Originally Posted by trigonometry424 yea sorry it was an accident i edited my op and would x = 1/2? No. but not x. It's probably easier to start with when and then - delete as appropriate depending on your boundary conditions For we have as the principal solution but there is another one to find
ohh yea so the answer would be pi/6 and 5pi/6 right?
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