How do you solve the following equation # sin x = 0.25# in the interval [0, 2pi]?

1 Answer
Monzur R.
Oct 5, 2016

#x_1=0.25 and x_2 = 2.89#

Explanation:

The easiest way to solve this type of problem is to plot the #sin# function and #f(x) = 0.25# on one graph (link to this here) and look for the intersections.

The graph shows two intersections: the principal value of #x#, labelled in the answer as #x_1#, and the other value of #x#, labelled in the answer as #x_2#. To find the principal value of #x#, we simply use the inverse #sin# function, #sin^-1(x)#:

#x_1=sin^-1(0.25)=0.252680255142 approx 0.25 #

To find the other value of #x#, we subtract the principle value of #x# from #pi#:

#x_2=pi-0.252680255142=2.8889123984477932 approx 2.89#

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