How do you solve $cos^2x = 1 + sinx$ on the interval [0,2pi]?

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Answer 1 · 2016-03-24T00:21:53

$x= 0, pi, 2pi and (3pi)/2$

$cos^2 x = 1 - sin^2 x$ , hence the given equation would become $1-sin^2 x = 1+ sin x$

Or, $" "sin^2 x +sin x=0$

Or, $" "sin x (sinx +1)=0$

This give $sin x=0$ and $sin x=-1$

$x= 0, pi, 2pi and (3pi)/2$

How do you solve cos^2x = 1 + sinx cos^2x = 1 + sinx on the interval [0,2pi]?