Question

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

Answer 1

`x in {0,pi/2,pi}`

Explanation:

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.

Answer 2

The solutions are `x=0,pi`.

Explanation:

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`