hi i was wondering if anyone could help me with this question
solve the equation 4 cot^2 x+12cosecx+1=0 giving all values of x to the nearest degree in the interval 0<x<360(should be equal to signs as well but comp doesnt do them)
thanks for any help much appreciated
solve the equation 4 cot^2 x + 12cosecx + 1 = 0 for 0<x<360
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stapel_eliz
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A good way to start is often to convert everything to sines and cosines, and see where that leads.
. . . . .}{\sin^2(x)}\, +\, \frac{12}{\sin(x)}\, +\, 1\, =\, 0)
In this case, that technique doesn't look very helpful. Try a Pythagorean identity instead:
. . . . .\right)\, +\, 12\csc(x)\, +\, 1\, =\, 0)
. . . . .\, +\, 12\csc(x)\, +\, 5\, =\, 0)
This is a quadratic in cosecant. You can solve by factoring, and then solve the two resulting trigonometric equations.
. . . . .
In this case, that technique doesn't look very helpful. Try a Pythagorean identity instead:
. . . . .
. . . . .
This is a quadratic in cosecant. You can solve by factoring, and then solve the two resulting trigonometric equations.
Re: solve the equation 4 cot^2 x + 12cosecx + 1 = 0 for 0<x<360
hiya thanks for the answer not sure if you have worked it out or not but would the answer be 22.925 and 202.925 or not
thanks for any help
thanks for any help
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stapel_eliz
- Posts: 1628
- Joined: Mon Dec 08, 2008 4:22 pm
- Contact:
Re: solve the equation 4 cot^2 x + 12cosecx + 1 = 0 for 0<x<360
o right thank you didnt think of testing it lol