Question

2sinxcscx-cscx=4sinx-2.. find the solutions of the equation that are in the interval [0, 2pi)

Answer 1

\[2\sin x \frac{1}{\sin x} - \frac{1}{\sin x} = 4\sin x -2\]
\[2-\frac{1}{\sin x} = 4\sin x -2\]
\[-\frac{1}{\sin x} = 4\sin x -4\]
\[-1 = 4\sin^{2} x -4\sin x\]
\[4\sin^{2} x -4\sin x +1 = 0\]
factor
\[(2\sin x -1)^{2} = 0\]
\[2\sin x = 1\]
\[\sin x = \frac{1}{2}\]
\[\rightarrow x = \frac{\pi}{6}, \frac{5\pi}{6}\]