How do you find a numerical value of one trigonometric function of x given #cos^2x+2sinx-2=0#?

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Answer 1 · 2016-11-14T19:36:28

#:.x=pi/2 + 2pin#

#cos^2x+2sinx-2=0#

#(1-sin^2x)+2sinx-2=0#

#-sin^2x+2sinx-1=0#

#sin^2x-2sinx+1=0#

Now factor.

#(sinx-1)(sinx-1)=0#

#(sinx-1)^2=0#

#sinx-1=0#

#sinx=1#

#x=sin^-1 1#

#:.x=pi/2 + 2pin#