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

1 Answer

bp · Stefan V. · mason m

Mar 24, 2016

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

Explanation:

#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#

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