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

enter image source here

2 Answers
Mar 11, 2018

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.

Mar 11, 2018

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