How do you solve #sec theta = 4#?

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Answer 1 · 2018-05-26T16:42:27

#theta=-arccos(1/4)+2kpi# or #theta=arccos(1/4)+2kpi#

Since
#sec(x)=1/cos(x)#
we have to solve that
#cos(theta)=1/4#
We get the solutions like above.

Answer 2 · 2018-05-26T17:56:04

#t = +- 75^@52 + k360^@#

sec t = 4 --># cos t = 1/(sec t) = 1/4 = 0.25#
Calculator and unit circle give 2 solutions for t:
#t = +- 75^@52 + k360^@#