How do you solve #sinx+3=3#?

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Answer 1 · 2016-09-13T10:57:13

#x = 0#

We have: #sin(x) + 3 = 3#

First, let's subtract #3# from both sides of the equation:

#=> sin(x) + 3 - 3 = 3 - 3#

#=> sin(x) = 0#

Then, let's take the #arcsin# of both sides:

#=> arcsin(sin(x)) = arcsin(0)#

#=> x = 0#

Answer 2 · 2016-09-13T14:33:49

#x = kpi#

sin x + 3 = 3
sin x = 0
Unit circle -->
sin x = 0 --> x = 0, #x = pi#, and #x = 2pi#
General answers:
#x = kpi#