SOLUTION: I need help solving for 2sin(2x-π/6)+1=0 xϵ(-π,π)

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Question 1081707: I need help solving for 2sin(2x-π/6)+1=0 xϵ(-π,π)
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
2sin%282x-pi%2F6%29%2B1=0
2sin%282x-pi%2F6%29=-1
sin%282x-pi%2F6%29=-1%2F2
So,
2x-pi%2F6=%287%2F6%29pi%2B2pi%2Ak and 2x-pi%2F6=%2811%2F6%29pi%2B2pi%2Ak
2x=%288%2F6%29pi%2B2pi%2Ak and 2x=%2812%2F6%29pi%2B2pi%2Ak
x=%282%2F3%29pi%2Bpi%2Ak and x=pi%2Bpi%2Ak
and
x=%282%2F3%29pi
x=pi
x=%285%2F3%29pi
x=2pi
Adjusting for the range, subtract pi from each answer.
x=-%281%2F3%29pi
x=0
x=%282%2F3%29pi
x=pi
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