How do you solve #sec^5x = 4secx# from 0 to 2pi?

1 Answer
Nghi N.
Aug 13, 2015

Solve #sec^5 x = 4sec x#

Ans:# +- pi/4, +- (3pi)/4)#

Explanation:

#sec^5 x - 4sec x = 0#

#sec x(sec^4 x - 4) = sec x(sec^2 x - 2)(sec^2 x + 2) =#

#= sec x(sec x - sqrt2)(sec x + sqrt2)(sec^2 x + 2) = 0#

a. #sec x = 1/cos x = 0# (undefined)
b. #1/cos x = sqrt2# --> #cos x = 1/sqrt2 = sqrt2/2# --> #x = +- pi/4#
c. #sec x = 1/cos x = -sqrt2# --> #cos x = -1/sqrt2# --> #x = +-(3pi)/4#
d. (sec^2 x + 2) (always positive)

Answers: #+- pi/4, +- (3pi)/4#
Note. #Arc -pi/4# is arc #(7pi)/4#, and arc #-(3pi)/4# is the arc #(5pi)/4#

Related questions
See all questions in Solving Trigonometric Equations
Impact of this question
2319 views around the world
You can reuse this answer
Creative Commons License