How to find the value of x where $0<=x<=360^@$ for $sec x= 3.1909$ ?

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Answer 1 · 2015-05-31T02:43:34

$sec x = 1/cos x = 3.1909 -> cos x = 1/3.1909 = 0.31$

$cos x = 0.31 -> x = +- 71.94 deg$

Within period (0, 2pi):

x = 71.94 deg
x = 360 - 71.94 = 288.06 deg

How to find the value of x where 0<=x<=360^@0<=x<=360^@ for sec x= 3.1909sec x= 3.1909?