How can I solve the equation for this interval, (see picture)? Thanks!

Rewrite the equation as

#tan^2x(1-sin x)=0#

So, either #tan x = 0# or, #sin x = 1#.

In the interval #[0,2pi)#, there is only one value of #x# for which #sin x=1#. This is #x=pi/2#.

On the other hand, #tan x=0# can give us two solutions for #x# in this interval, namely #x=0# and #x=pi#.

Thus, the possible values of #x# in the interval #[0,2pi)# belongs to the set #{0,pi/2,pi}#

Note : None of the options given in the picture are correct (or, if you are not looking for all solutions, the lower three are)!
While #x = 2pi# does satisfy the equation - it does not belong to the interval #[0,2pi)# - so the first option is definitely wrong.

Treat the problem like a quadratic; factor it, then set the factors equal to #0#:

#tan^2xsinx=tan^2x#

#tan^2xsinx-tan^2x=0#

#color(blue)(tan^2x)*sinx-color(blue)(tan^2x)*1=0#

#color(blue)(tan^2x)(sinx-1)=0#

Now set each of the factors equal to zero:

#color(white){color(black)( (tan^2x=0, qquadqquad sinx-1=0), (tanx=0,qquadqquad sinx=1), (x=0 and pi,qquadqquad x=pi/2):}#

Since #pi/2# doesn't work when you plug it into the original equation (because #tancolor(black)(pi/2)# is undefined), it is not a solution.

The other two work, so they are real solutions to the problem.

#x=0,pi#