Given the equation:
[tex]3\sin ^2x\cot x-3\cot x=0[/tex]Let's solve the given equation over the interval:
[tex]\lbrack0^o,360^o)[/tex]Add 3cotx to both sides of the equation:
[tex]\begin{gathered} 3\sin ^2x\cot x-3\cot x+3\cot x=0+3\cot x \\ \\ 3\sin ^2x\cot x=3\cot x \end{gathered}[/tex]Cancel the common factors:
[tex]3\sin ^2x=3[/tex]Divide both sides by 3:
[tex]\begin{gathered} \frac{3\sin ^2x}{3}=\frac{3}{3} \\ \\ \sin ^2x=1 \end{gathered}[/tex]Take the square root of both sides:
[tex]\begin{gathered} \sin x=\sqrt[]{1} \\ \\ \sin x=1 \end{gathered}[/tex]Take the sine inverse of both sides:
[tex]\begin{gathered} x=\sin ^{-1}(1) \\ \\ x=90^o \end{gathered}[/tex]Now, the sine function is positive in quadrant I and II, to find the second solution, add 180 to the reference angle.
x = 90 + 180 = 270 degrees.
Therefore, the solutions to the equation on the given interval:
x = 90⁰, 270⁰
ANSWER:
x = 90⁰, 270⁰