Assuming a restriction of x in the interval [0, 2pi):
sqrt(3) (tan4x) + 1 = 0
Move the 1 over to the right hand side,
sqrt(3) tan4x = -1
Divide both sides by sqrt(3),
tan(4x) = -1/sqrt(3)
Now, where on the graph is tan equal to -1/sqrt(3)? We at least know it's going to be in quadrants 2 and 4. The answer is at 5pi/6 and 11pi/6
Therefore,
4x = {5pi/6, 11pi/6, ?, ?, ?, ?, ?, ?}
Since we have that 4x, we have to quadruple the number of solutions, hence the question marks. What we're going to have to fill the question marks are
5pi/6 + 2pi, 5pi/6 + 4pi, 5pi/6 + 6pi
11pi/6 + 2pi, 11pi/6 + 4pi, 11pi/6 + 6pi
4x = {5pi/6, 11pi/6, 17pi/6, 23pi/6, 29pi/6, 35pi/6, 41pi/6, 47pi/6}
Now, we multiply both sides by 1/4, which means all of the denominators get multiplied by 4.
x = {5pi/24, 11pi/24, 17pi/24, 23pi/24, 29pi/24, 35pi/24, 41pi/24, 47pi/24}
2007-01-09 16:41:06
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answer #2
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answered by Puggy 7
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(â3) tan(4x) + 1 = 0
tan(4x) = -1/â3
4x = arctan(-1/â3) = 5Ï/6 + Ïn, where n is an interger
x = (1/4)(5Ï/6 + Ïn) = 5Ï/24 + Ïn/4 = 5Ï/24 + 6Ïn/24
2007-01-09 17:33:55
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answer #3
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answered by Northstar 7
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if by this you mean
sqrt(3)(tan(4x + 1)) = 0
tan(4x + 1) = 0
4x + 1 = 0
4x = -1
x = (-1/4)
However if you mean
sqrt(3)tan(4x) + 1 = 0
sqrt(3)tan(4x) = -1
tan(4x) = -1/(sqrt(3))
tan(4x) = (-sqrt(3)/3)
4x = 330 or 150
x = 82.5 or 37.5
2007-01-09 17:16:00
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answer #4
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answered by Sherman81 6
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tan4x= -1/sqrt3 = tan(-pie/6)
now if
tan a=tan b
a=n*pie +b , n = an integer;
thus 4x=(n*pie - pie/6)
=> x = (1/4)(n*pie - pie/6)
2007-01-09 16:41:45
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answer #5
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answered by Rahul 2
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