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FInd all the angles between 0 and 2 pi which satisfy the following equations.
3 cosec (2x - 3) +15 =1
managed to find that x = 3.18 and 4.53. however i am still missing 2 more answers and that is 0.0372 and 1.39. the range is -3 < 2x - 3 < 4 pi - 3
May I ask if there's anyone who can help assist me in this question? i tried to look at other viabilities to obtaining these 2 final answers but couldn't seem to get them. thanks.
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0 < x < 2pi
0 < 2x < 4pi
-3 < 2x - 3 < 4pi - 33 cosec (2x - 3) + 15 = 1
cosec (2x - 3) = -14/3
sin (2x - 3) = -3/14basic angle = 0.21596
2x - 3 = (-pi + 0.21596) , (0 - 0.21596) , (pi + 0.21596) , (2pi - 0.21596)
x = 0.0372 , 1.39 , 3.18 , 4.53since the range goes beyond 0 to the negative region, u have to consider the angles in that region too
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Originally posted by wishboy:
0 < x < 2pi
0 < 2x < 4pi
-3 < 2x - 3 < 4pi - 33 cosec (2x - 3) + 15 = 1
cosec (2x - 3) = -14/3
sin (2x - 3) = -3/14basic angle = 0.21596
2x - 3 = (-pi + 0.21596) , (0 - 0.21596) , (pi + 0.21596) , (2pi - 0.21596)
x = 0.0372 , 1.39 , 3.18 , 4.53since the range goes beyond 0 to the negative region, u have to consider the angles in that region too
yea, i get what you are saying. what i do not know is why it is (-pi + 0.21596) and (0-0.21596)...could you tell me how you know how to go about you doing the respective angles? why is it ( -pi +..., and 0 -....) i haven't dealt with negative regions so far so i don't kind of get the idea.
thanks! (=
Edited by bonkysleuth 26 Aug `08, 9:49PM
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2nd | 1st
-------+-------
3rd | 4thdun really know how to explain...
the above shows the 4 different quadrants
for positive region, the angle starts from 1st --> 2nd --> 3rd and so on, in an anticlockwise manner
for negative region, the angle starts from 4th --> 3rd --> 2nd and so on, in a clockwise manner-3 goes past the 4th and 3rd quadrants
as sin (2x - 3) gives a negative value, x is in the 3rd and 4th quadrants
so u would have to use (-pi +..., and 0 -....)
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