How do you solve $sin x + 3 = 3$ $sinx+3=3$ ?

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Answer 1 · 2016-09-13T10:57:13

$x = 0$ $x = 0$

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

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

$\Rightarrow sin (x) + 3 - 3 = 3 - 3$ $=> sin(x) + 3 - 3 = 3 - 3$

$\Rightarrow sin (x) = 0$ $=> sin(x) = 0$

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

$\Rightarrow arcsin (sin (x)) = arcsin (0)$ $=> arcsin(sin(x)) = arcsin(0)$

$\Rightarrow x = 0$ $=> x = 0$

Answer 2 · 2016-09-13T14:33:49

$x = k π$ $x = kpi$

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

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