SOLUTION: Find solutions in radians for 0 less then or equal to x less then 2pi: sin^2(3x)-3sin(3x)-4=0

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Question 744248: Find solutions in radians for 0 less then or equal to x less then 2pi:

sin^2(3x)-3sin(3x)-4=0
Found 2 solutions by lwsshak3, Alan3354:
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Find solutions in radians for 0 less then or equal to x less then 2pi:
sin^2(3x)-3sin(3x)-4=0
sin((3x)-4)(sin(3x)+1)=0
..
sin((3x)-4)=0
sin(3x)=4 (reject,(-1 ≤ sin(3x) ≤ 1))
..
sin(3x)+1=0
sin(3x)=-1
3x=3π/2
x=3π/6=π/2

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Find solutions in radians for 0 less then [sic] or equal to x less then 2pi:
---
It's "less than", not less then
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Say it like this: 0 <= x <= 2pi
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sin^2(3x)-3sin(3x)-4=0
Sub u for sin(3x)
u^2 - 3u - 4 = 0
(u - 4)*(u + 1) = 0
u = 4 --> no real # for arcsin(4)
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u = -1 --> sin(3x) = -1
..
sin(3x)=-1
3x = 3pi/2, 7pi/2, 11pi/2
x = pi/2, 7pi/6, 11pi/6