Question

The equation $2{\sin }^3\theta +(2K-3){\sin }^2\theta -(3K+2)\sin \theta -2K=0$ has exactly three roots in $(0,2\pi )$ then $K$ can be equal to

• A. $0$
• B. $\frac{1}{2}$
• C. $1$
• D. $-1$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $0$
C $1$
D $-1$