How do you solve #1+secx=2cosx#?

1 Answer
Dec 30, 2016

#0, (2pi)/3, (4pi)/3, 2pi#

Explanation:

1 + sec x = 2cos x
#1 + 1/(cos x) = 2cos x#
#cos x + 1 = 2cos^2 x#
#2cos^2 x - cos x - 1 = 0#
Solve this quadratic equation for cos x.
Use trig table and unit circle.
Since a + b + c = 0, use shortcut. There are 2 real roots:
cos x = 1 and cos x = c/a = - 1/2.
a. cos x = 1 --> x = 0 and #x = 2pi#
b. #cos x = - 1/2# -->arc # x = +- (2pi)/3#
Co-terminal of arc #(-2pi)/3# is arc #(4pi)/3#
Answers for #(0, 2pi)#:
#0, (2pi)/3, (4pi)/3, 2pi#
For general answers add #2kpi#