How do you find the solution to #sec^2theta-7sectheta+12=0# if #0<=theta<360#?

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Answer 1 · 2017-01-11T14:27:44

#x=70.53° or x=75.52° or x=289.47 or x=284.48°#

You would use the quadratic formula to solve #secx#

#secx=(7+-sqrt(49-48))/2=(7+-1)/2#

#secx=3 or secx=4#

Since

#secx=1/cosx#

you would solve:

#1/cosx=3 or 1/cosx=4#

that's

#cosx=1/3 or cosx=1/4#

#x=arccos(1/3) or x=360-arccos(1/3) or x=arccos(1/4) or x=360-arccos(1/4)#

#x=70.53° or x=75.52° or x=289.47 or x=284.48°#