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

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

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