SOLUTION: Solve each equation for solutions over the interval [0,2pi). 2tanx - 1 = 0 I did x = tan^-1(1/2)but I did not get the right answer. Correct answer: {0.463647609,3.605240

Algebra ->  Trigonometry-basics -> SOLUTION: Solve each equation for solutions over the interval [0,2pi). 2tanx - 1 = 0 I did x = tan^-1(1/2)but I did not get the right answer. Correct answer: {0.463647609,3.605240      Log On


   

Question 809378: Solve each equation for solutions over the interval [0,2pi).
2tanx - 1 = 0
I did x = tan^-1(1/2)but I did not get the right answer.
Correct answer: {0.463647609,3.605240263}
Could you please explain how to solve this and these types of problems? I am having a really hard time understanding this.
Thanks!
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
2%2Atan%28x%29-+1+=+0
2%2Atan%28x%29=1
tan%28x%29=1%2F2
The calculator would tell you that tan%5E%28-1%29%280.5%29=0.463647609 in radians,
or tan%5E%28-1%29%280.5%29=26.56505118%5Eo in degrees.
That is because the inverse function of tangent is defined as the angle that has that tangent and is between -pi%2F2 and pi%2F2 in radians,
or between -90%5Eo and 90%5Eo in degrees.
However, the function tangent has a period of pi radians (or 180%5Eo in degrees.
That means, in other words, that tan%28x%2Bpi%29=tan%28x%29 .
That gives you the other solution,
x=0.463647609%2Bpi=3.605240263