## Shaydenparis solve the following equation on interval [0,2pi) 1-cosx=-sinx one year ago one year ago

Let's write c instead of $$\cos x$$ and s instead of $$\sin x$$. This is just to reduce the amount of writing involved. $1-c = -s$Square both sides $(1-c)(1-c) = 1-2c+c^2 = s^2$Remember that $\sin^2x+\cos^2x = 1$so we can rewrite the right hand side as $1-2c+c^2=1-c^2$If we solve that for $$c$$ $-2c+2c^2 = 0$$-c+c^2=0$$c^2=c$$c=1$And now we undo our substitution to reveal$\cos x = 1$Where does $$\cos x=1$$ in the interval $$[0,2\pi)$$?