How do you find all solutions to #sin 2x= 0.2# on [0,2pi)?

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Answer 1 · 2016-04-18T17:03:32

#S={0.1007,1.4701,3.2423, 4.6117}#

#2x=sin^-1 0.2#

Note that since sine is positive in Quadrants I and II we have two solutions

#2x=0.2013....+2pin, pi-0.2013...+2pin=2.9402...+2pin#

#x=0.10067...+pin,1.4701...+pin#

#n=0, x=0.1007,1.4701#

#n=1, x=3.2423, 4.6117#

#n=2, x=6.3839, 7.7533->#this is greater than #2pi #so we stop

#S={0.1007,1.4701,3.2423, 4.6117}#