Question

Solve the equation on the interval [0, 2π).

tan^2 x sin x = tan^2 x
B. pi/2, 2π
C. pi/2, π
D. 0, π

Answer 1

@studystagram

Answer 2

I got D. is that correct

Answer 3

\[tan^2(x)sin(x)=tan^2(x)\]
\[tan^2(x)sin(x)-tan^2(x)\]
\[tan^2(x)*(sin(x)-1)=0\]Therefore
\[tan^2(x)=0~~or~~sin(x)-1=0\]
if\[x=\pi\\tan^2(\pi)=0\]
if\[x=\frac{\pi}{2}\\sin\left(\frac{\pi}{2}\right)=1\]
but, if\[x=0\\tan^2(0)=0\]The solutions of this equation will be\[S=\left\{0,\frac{\pi}{2},\pi\right\}\]