Trigonometric equation sqrt(2)sin(x/3)+1=0, 2 solutions are 15pi/4 + 6pi(k) and 21pi/4 + 6pi(k).... why

Find all solutions of the equation. (Enter each answer in the form θ + aπk, with θ ≥ 0 and θ as small as possible.


my steps:

sqrt(2)sin(x/3)+1=0

substitute y for x/3

sin(y) = -1/sqrt2 <-- both positive and negative quadrants

sin^-1(y) = sin^-1(sqrt2/2)

y = pi/4

x/3 = pi/4

x = 3pi/4.....

How can I get from 3pi/4 to:

15pi/4 + 6pi(k) and 21pi/4 + 6pi(k)??

thanks!

√2∙sin(x/3) + 1 = 0

sin(x/3) = -1/√2 = -√(2)/2

x/3 = 5π/4 + 2nπ, 7π/4 + 2nπ

x = 15π/4 + 6nπ, 21π/4 + 6nπ

Using inverse trig functions will only give you one solution (due to the fact that they are "functions") so use your unit circle.

Without substitution, you get:

sin(x/3) = -1/sqrt(2) = -sqrt(2)/2

When solving a trigonometric equation, most of the time you'll be able to get a solution off of the unit circle, rather than having to use the inverse trig function (i.e. sin^-1(x)).

Here, we can, as x/3 = 5pi/4 and 7pi/4.

Add to these solutions 2pi, since sine is periodic.

x/3 = 5pi/4 + 2pi*n and 7pi/4 + 2pi*n

x = 15pi/4 + 6pi*n and 21pi/4 + 6pi*n

Your mistake was solving for the angle incorrectly.

x = 3π/4 is wrong.

(√2)sin(x/3) + 1 = 0
(√2)sin(x/3) = -1
sin(x/3) = -1/√2
sin(x/3) = -(√2)/2
If you know your unit circle, you'll know that
x/3 ∈ {5π/4 + 2πk,7π/4 + 2πk}, where k ∈ Z.
So x ∈ {15π/4 + 6πk,21π/4 + 6πk}.