SOLUTION: Any help would be much appreciated in figure out this problem. What are all the solutions in the equation tan^2 x + tanx=0 in the interval [0,2pi). Thank you!

Algebra ->  Trigonometry-basics -> SOLUTION: Any help would be much appreciated in figure out this problem. What are all the solutions in the equation tan^2 x + tanx=0 in the interval [0,2pi). Thank you!      Log On


   

Question 1206427: Any help would be much appreciated in figure out this problem. What are all the solutions in the equation tan^2 x + tanx=0 in the interval [0,2pi).
Thank you!
Found 3 solutions by MathLover1, math_tutor2020, ikleyn:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

tan%5E2+%28x%29+%2B+tan%28x%29=0+
tan+%28x%29+%28tan%28x%29%2B1%29=0+
solutions:
tan%28x%29=0
or
tan%28x%29=-1

then
x=tan%5E-1%280%29
x=0
or
x=tan%5E-1%28-1%29
x=-pi%2F4

in the interval [0,2pi)
a function tan%28x%29 is a periodic function and has a period of +pi
x=0
x=0%2Bpi=pi
x=-pi%2F4%2Bpi=3pi%2F4
x=3pi%2F4%2Bpi=7pi%2F4


Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Answers:
x+=+0, x+=+3pi%2F4, x+=+pi, x+=+7pi%2F4

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Work Shown

tan%5E2%28x%29+%2B+tan%28x%29+=+0

tan%28x%29%28tan%28x%29+%2B+1%29+=+0

tan%28x%29=0 or tan%28x%29+%2B+1+=+0

tan%28x%29=0 or tan%28x%29+=+-1

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tan%28x%29+=+0

sin%28x%29%2Fcos%28x%29+=+0

sin%28x%29+=+cos%28x%29%2A0

sin%28x%29+=+0

This leads to x+=+0 and x+=+pi when considering the interval [0, 2pi)
Refer to the unit circle. These angles occur when the y coordinate is 0.

We exclude x = 2pi since [0, 2pi) means 0+%3C=+x+%3C+2pi

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tan%28x%29+=+-1

sin%28x%29%2Fcos%28x%29+=+-1

sin%28x%29+=+-cos%28x%29

Look on the unit circle for terminal points where x = -y or y = -x.
In other words, look for points on the unit circle that have the same coordinates just one is negative.

Tangent is negative in quadrants II and IV

The solution in quadrant II is x+=+3pi%2F4 and the solution in quadrant IV is 7pi%2F4 (they are separated by a gap of pi radians aka 180 degrees).

Answer by ikleyn(52794) About Me  (Show Source):
You can put this solution on YOUR website!
.

Actually, tan(x) has a basic period from 0 to pi.


The interval from 0 to 2pi  is TWO basic periods of tan(x).


Therefore, to simplify the problem, you could find the solutions to the given equation in the interval [0,pi) first,
and then to extend (to translate) the roots from this interval [0,pi) to one interval forward.


In the basic interval [0,pi), the roots are 0 and 3pi%2F4.


After moving forward one period, two other roots are added to the set of solutions,  pi  and  7pi%2F4.

Solved.


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Generally speaking,  in  Math it is not a good practice / (is not a good style)
to ask about the solutions of the equation

                f(x) = 0,

where  f(x)  is periodical function with the period  T,  in two-period interval  [0,2T).

It is not a good style.

Such questions,  as a rule,  go about one single period interval  [0,T).


Otherwise,  questions about mathematical competence of the author may arise.