How do you find the exact value of #sec^2theta-10sectheta+20=0# in the interval #0<=theta<2pi#?

1 Answer
Jul 26, 2017

#t = +- 82^@06#
#t = +- 68^@90#

Explanation:

#sec^2 - 10 sec + 20 = 0#
Solve this quadratic equation for sec x by using the quadratic formula in graphic form (Google Search):
#D = d^2 = b^2 - 4ac = 100 - 80 = 20# --> #d = +- 2sqrt5#
There are 2 real roots:
#sec t = -b/(2a) +- d/(2a) = 10/2 +- 2sqrt5/2 = 5 +- sqrt5#
Two values of cos t
#cos t = 1/(5 + sqrt5)# and #cos t = 1/(5 - sqrt5)#
a. #cos t = 1/7.24 = 0.138#
Calculator and unit circle give:
#t = +- 82^@06#
b. #cos t = 1/(5 - sqrt5) = 1/2.76 = 0.36#
#t = +- 68^@90#