SOLUTION: Solve for x on the interval [0,2pi]. tan^2(x)-1=0 Someone please help, this assignment is due tomorrow and I cannot find how to do this anywhere.

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Question 1043019: Solve for x on the interval [0,2pi].
tan^2(x)-1=0
Someone please help, this assignment is due tomorrow and I cannot find how to do this anywhere.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
tan^2(x)-1=0

(tan(x)+1)(tan(x)-1)=0 ... difference of squares rule

tan(x)+1=0 or tan(x)-1=0

Let's focus on each sub-equation one at a time.

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Let's solve the first sub-equation for x.

tan(x)+1=0

tan(x) = -1

x = arctan(-1)

x = 3pi/4 or x = 7pi/4

Note: use the unit circle to determine the two angles. Look at where the x and y coordinates are equal and opposite in sign (Quadrant 2 and 4)

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Now move onto the next

tan(x)-1 = 0

tan(x) = 1

x = arctan(1)

x = pi/4 or x = 5pi/4

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The four solutions are:
x = pi/4
x = 3pi/4
x = 5pi/4
x = 7pi/4