Question

Solve the equation on the interval [0, 2π). tan^2x sin x = tan^2x

Answer 1

Start by getting everything on the same side of the equals sign and then set it equal to 0.  $tan^2xsinx-tan^2x=0$.  Factor out the common tan^2x like this: $tan^2x(sinx-1)=0$.  Now we have 2 separate equations to solve: $tan^2x=0$ and sinx = 0.  Now we have to figure out where tan^2 is 0 between 0 and 2pi.  If we include 2pi, the solutions for that equation are $x = 0, \pi , 2 \pi$.  You can test those out on your calculator just to be sure.  There's only one value of x for the next equation.  The only place between 0 and 2pi where the sin x = 1 is at x = $\frac{ \pi }{2}$.  And there you go!