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

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