SOLUTION: solve the equation on the interval [0,2pi] tan^2(x)=tan^2(x)sec(x)

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Question 1024815: solve the equation on the interval [0,2pi]
tan^2(x)=tan^2(x)sec(x)
Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
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solve the equation on the interval [0,2pi]
tan^2(x)=tan^2(x)sec(x)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

tan%5E2%28x%29 = tan%5E2%28x%29%2Asec%28x%29  --->

tan%5E2%28x%29 - tan%5E2%28x%29%2Asec%28x%29 = 0,

tan%5E2%28x%29%2A%281-sec%28x%29%29 = 0.


So this equations breaks down into two independent equations


1.  tan%5E2%28x%29 = 0  --->  tan(x) = 0  --->  x = 0 and x = pi.


2.  1-sec%28x%29 = 0  ---> 1+-+1%2Fcos%28x%29 = 0  --->  cos(x) = 1  --->  x = 0. 


The entire set of solution of the original equation in the given interval consists of two numbers:  0  and  pi.