How do you solve #sin^2x + 2cosx = 2# over the interval 0 to 2pi?

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Answer 1 · 2018-03-02T09:05:25

#color(blue)(x=0,2pi#

Identity:

#color(red)bb(sin^2x+cos^2x=1)#

#sin^2x+2sinx=2#

Using identity:

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

Rearranging, simplifying and equating to zero:

#cos^2x-2cosx+1=0#

This is just a quadratic in #cosx#

Let #cosx=u#

#u^2-2u+1#

Factor:

#(u-1)(u-1)=0=>u=1#

#u=cosx#

#cosx=1#

#x=arccos(cosx)=arccos(1)=>color(blue)(x=0,2pi#