How do you solve sec #x# = cosec #3/4pi# for the range value of #0 <= x <= 2pi# ?

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Answer 1 · 2018-04-01T01:59:07

#x=pi/4, (7pi)/4#

First, let's evaluate #csc((3pi)/4).# Knowing that #cscx=1/sinx, csc((3pi)/4)=1/sin((3pi)/4)#

Now, #sin((3pi)/4)=sqrt(2)/2, csc((3pi)/4)=1/(sqrt(2)/2)=2/sqrt(2)#

So, we are really just being asked to solve

#secx=2/sqrt(2)#

Recalling that #secx=1/cosx:#

#1/cosx=2/sqrt(2)#

Invert both sides to get the cosine on top:

#cosx=sqrt(2)/2#

In the interval #[0, 2pi],# this holds true for #x=pi/4, (7pi)/4#