Thread: Solving trig equation

1]Show that 2[sinxcosx - 1/2 ] = sin2x -1

2] Hence, determine the general solution for:

2[sinxcosx - 1/2 ] =0

$\displaystyle 2(sinxcosx-\frac{1}{2}) = 2sinxcosx-2*\frac{1}{2} = 2sinxcosx-1 $ We know that $\displaystyle sin2x=2sinxcosx$ so we have
$\displaystyle 2sinxcosx-1 = sin2x-1$

thanks alot!