SOLUTION: Solve the equation on the interval [0, 2pi). tan^2 x sin x = tan^2 x B.pi/2 , pi D. 0, pi

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Question 1074774: Solve the equation on the interval [0, 2pi).
tan^2 x sin x = tan^2 x
B.pi/2 , pi
D. 0, pi
Answer by ikleyn(52780) About Me  (Show Source):
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Solve the equation on the interval [0, 2pi).
tan^2 x sin x = tan^2 x
B.pi/2 , pi
D. 0, pi
~~~~~~~~~~~~~~~~~~~~ 

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

tan%5E2%28X%29%2A%28sin%28x%29-1%29 = 0.


This equation deploys in two independent equations:


1.  tan(x) = 0 with the solutions x = 0 and/or x = pi.


2.  sin(x) = 1 with the solution  x = pi%2F2.

    This solution is EXTRANEOUS, since  tan is not defined for x = pi%2F2.


Answer.  The solutions are  x = 0  and  x = pi.