Originally Posted by
Jhevon no.
$\displaystyle 1 - \cos^2 x = \sin^2 x$
you have $\displaystyle 1 - \cos x$ here, you cannot apply that identity.
we have: $\displaystyle \frac {2 \sin x + 1 - \cos x }{\cos x } = 0$
$\displaystyle \Rightarrow 2 \sin x + 1 - \cos x = 0$ .....(a fraction is zero when the numerator is zero, provided the denominator is not zero at the same time)
$\displaystyle \Rightarrow 2 \sin x + 1 = \cos x$
$\displaystyle \Rightarrow \left( 2 \sin x + 1 \right)^2 = \cos^2 x$
$\displaystyle \Rightarrow \left( 2 \sin x + 1 \right)^2 = 1 - \sin^2 x$
now can you continue?