sin2x+sinx-2cox-1=0
As we know, Sin 2x =2sinx cosx
2Sinx Cosx+Sinx-2cosx-1=0
Sinx(2cos x+1)-1(2cosx+1)=0
(sinx-1)(2cosx+1)=0
So, either sinx-1=0 or 2cos x+1=0
Sin x=1 or cos x=[tex] \frac{-1}{2} [/tex]
x=[tex] \frac{\pi}{2} [/tex] or multiples of [tex] \frac{\pi}{2} [/tex]
x=[tex] \frac{2\pi}{3} [/tex] or multiples of [tex] \frac{2\pi}{3} [/tex]
x=[tex] \frac{-2\pi}{3} [/tex] or multiples of [tex] \frac{-2\pi}{3} [/tex]
