How do you solve #1+sinx=2cos^2x# and find all exact general solutions?

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Answer 1 · 2016-11-07T17:14:22

Use the identity #sin^2theta + cos^2theta = 1 -> cos^2theta = 1- sin^2theta#.

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

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

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

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

#2sinx(sinx + 1) - (sinx + 1) =0#

#(2sinx - 1)(sinx + 1) = 0#

#sinx= 1/2 and sinx = -1#

#x= pi/6 + 2pin, (5pi)/6 + 2pin, (3pi)/2 + 2pin, "where "#n#" is an integer"#

Hopefully this helps!