How do you solve #cosxtanx-2cos^2x=-1#?

1 Answer
Nghi N
Dec 15, 2016

#pi/6 + 2kpi#
#(5pi)/6 + 2kpi#

Explanation:

#cos x(sin x/(cos x)) = 2cos^2 x - 1#
Simplify by cos x. We get, with condition: cos x diff. to zero, or x diff. to #pi/2 and (3pi)/2#
#sin x = 2cos^2 x - 1 = cos 2x = 1 - 2sin^2 x #
Reminder of trig identity: #cos 2x = 2cos^2 x - 1 = 1 - 2sin^2 x#
Solve this quadratic equation for sin x:
#2sin^2 x + sin x - 1 = 0#
Since a - b + c = 0, use shortcut. The 2 real roots are:
sin x = - 1 and #sin x = - c/a = 1/2#
Use trig table and unit circle -->
a. sin x = - 1 --> arc #x = (3pi)/2#
This solution is rejected because of the above condition.
b. #sin x = 1/2# -->
#x = pi/6# and #x = pi - pi/6 = (5pi)/6#
General answers:
# pi/6 + 2kpi#
#(5pi)/6 + 2kpi#

Related questions
See all questions in Solving Trigonometric Equations
Impact of this question
6247 views around the world
You can reuse this answer
Creative Commons License