How do you solve the equation on the interval [0,2pi) for ?? 3sin theta-1=0

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Answer 1 · 2018-04-01T15:44:24

#color(blue)(theta=arcsin(1/3), pi-arcsin(1/3) #

#3sin(theta)-1=0#

#3sin(theta)=1#

#sin(theta)=1/3#

#theta=arcsin(sin(theta))=arcsin(1/3)=>#

#color(blue)(theta=arcsin(1/3), pi-arcsin(1/3) #

Approximate values:

#theta=0.3398 , 2.8017# radians