How do you solve 5=sec^2x+3 for 0<=x<=2pi?

1 Answer
Sep 12, 2016

x=pi/4, (3pi)/4, (5pi)/4, (7pi)/4

Explanation:

5=sec^2x + 3

sec^2x = 2

secx =+-sqrt(2)

cosx =+-1/sqrt(2)

Taking positive result:
x= arccos(1/sqrt(2)) -> x=pi/4 or (7pi)/4 for x in {0, 2pi)

Taking negative result:
x=arccos(-1/sqrt(2)) -> x= (3pi)/4 or (5pi)/4 for x in {0, 2pi)

Hence: x=pi/4, (3pi)/4, (5pi)/4, (7pi)/4 for x in {0, 2pi)