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Question Number 76439 by mathocean1 last updated on 27/Dec/19
solve in [0;2π[ 1+2sin3x≤0
solvein[0;2π[1+2sin3x⩽0
Answered by john santu last updated on 27/Dec/19
sin3x=−(1/2).→x={10^o ,50^o ,130^o ,170^o ,250^o ,290^o } therefore : 0^o ≤x≤10^o ∪50^o ≤x≤130^o ∪170^o ≤x≤250^o ∪290^o ≤x≤360^o
sin3x=−12.→x={10o,50o,130o,170o,250o,290o}therefore:0o⩽x⩽10o∪50o⩽x⩽130o∪170o⩽x⩽250o∪290o⩽x⩽360o

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