How do you solve #1 - sin^2x +2sinx = 2#?

1 Answer
Jan 22, 2018

#x = pi/2+ 2pin#

Explanation:

First of all, this is an equation, not an identity.

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

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

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

#sinx= 1#

#x = pi/2 + 2pin#

Note that I added the #2pin# because we know that the period of #sinx# is #2pi#.

Hopefully this helps!