Ask Me Help Desk - Trigonometric Equations

Quote:

Determine the solutions for (1+secx) / (secx) = (sin^2x) / (1-cosx) in the interbal x E (-2pi, 2pi)

$\frac {1+sec(x)}{sec(x)}=\frac {sin^2(x)}{(1-cos(x))}$

$\frac {1+sec(x)}{sec(x)}=\frac {(1-cos(x))(1+cos(x))}{(1-cos(x))}$

$\frac {1+sec(x)}{sec(x)}=(1+cos(x))$

$\frac {1+\frac {1}{cos(x)}}{\frac {1}{cos(x)}}=(1+cos(x))$

$\frac {(1+cos(x))}{cos^2(x)}=(1+cos(x))$

$\frac {1}{cos^2(x)}=1$

$1=cos^2(x)$

$\sqrt {1}=\pm 1=\sqrt {cos^2(x)}=cos(x)$

$arccos(\pm 1)=x$ in the interval x E (-2pi, 2pi) which is true when

$x=-2\pi,\,-\pi,\,0,\, \pi,\, 2\pi$ on the specified interval.