How do you find the values of #theta# given #sectheta=sqrt2#?

1 Answer
Binayaka C.
Feb 11, 2017

General solution : # theta = (2 n pi + pi/4) , theta = (2 n pi + (7pi)/4) # , where n is an integer # [n = 0, 1, 2,3 ,.........]#

Explanation:

#sec theta = sqrt 2 :. cos theta = 1/sqrt 2 ; cos 45^0 = 1/sqrt 2 , cos (360-45) =cos 315= 1/sqrt 2 :. theta = 45^0 (pi/4), 315^0((7pi)/4) # Note: #cos theta# is positive in 1st and last quadrant.

General solution : # theta = (2 n pi + pi/4) , theta = (2 n pi + (7pi)/4) # where n is an integer # [n = 0, 1, 2,3 ,.........]# [Ans]

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