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
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{{{tan^2(x)*sin(x)}}} = {{{tan^2(x)}}}  --->

{{{tan^2(X)*(sin(x)-1)}}} = 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/2}}}.

    This solution is EXTRANEOUS, since  tan is not defined for x = {{{pi/2}}}.


<U>Answer</U>.  The solutions are  x = 0  and  x = {{{pi}}}.
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