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I know that a principal solution of a trignometric equation is 2 values lying between
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hi
See pictures.
Because the tan function keeps repeating, you can get the same value for tan x every pi radians.
So if pi/3 is a solution, so is pi/3 + pi = 4pi/3
For sec 2
sec is 1/cos.
Cosine is +ve in the 1st quadrant (0, pi/2) and 4th quadrant (3pi/2, 2pi)
So once you have done inverse sec 2 and got the first angle, subtract from 2pi to get another.
http://www.mathsisfun.com/algebra/trig- … raphs.html
and
http://www.mathsisfun.com/algebra/trig- … rants.html
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Thanks
And if
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Hi mttal24
-ve means negative and +ve positive.
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sec x = -2
=> 1/cos x = -2
=> cos x = -1/2
cosine is negative between pi/2 and 3pi/2
note also: cos x = +1/2 has a solution pi/3
So the solutions we want (in the 2nd and 3rd quadrant) are pi - pi/3 = 2pi/3 and pi + pi/3 = 4pi/3
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Thanks.
Lets say, I begin answering by:
And, in some books I see that, suddenly out of the blues, a person adds
to and makes it
Now, I have understood that after reading the same question for the past 5 days.
But your method seems to be faster.
Last edited by mttal24 (2012-05-13 20:19:27)
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