How do you solve #1 - sin^2x +2sinx = 2#?
1 Answer
Jan 22, 2018
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
Hopefully this helps!