Question

solve the following equation on interval [0,2pi) 1-cosx=-sinx

Answer 1

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 \([0,2\pi)\)?

Answer 2

thank you so much