How to solve 3cosx+3=sin^2x over the interval (0,2pi]?

Question

#3cosx+3=sin^2x# Solve over the interval #(0,2pi]# I got #pi# as the only solution but can someone check to make sure that I am right? Thanks!

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Answer 1 · 2018-02-13T00:15:20

#x=pi#

.

#3cosx+3=sin^2x#

#3cosx+3=1-cos^2x#

#cos^2x+3cosx+2=0#

Let's factor it:

#(cosx+2)(cosx+1)=0#

#cosx+2=0 :. cosx=-2# This is not valid because #cos# values can only be between #1# and #-1#

#cosx+1=0#

#cosx=-1 :. x=pi#