How do you solve #cosx tanx - 2 cos^2x = -1# on the interval #[0,2pi]#?

1 Answer
Nghi N.
Jul 25, 2015

Solve: cos x.tan x - 2cos^2 x = - 1

Explanation:

#cos x(sin x /cos x) = 2cos^2 x - 1#
sin x = cos 2x (Condition cos x not zero, x not #(pi)/2# and #(3pi)/2#)
#sin x = 1 - 2sin^2 x#. Call sin x = t. We get quadratic equation:

#2t^2 + t - 1 = 0#
(a - b + c) = 0 --> 2 real roots: t = -1 and t = -c/a = 1/2

a. t = sin x = - 1 --> #x = (3pi)/2 # (Rejected)

b. #t = sin x = 1/2# --> #x = (pi)/6 #and #x = (5pi)/6#
Check.
#x = (pi)/6# --> #cos x = sqrt3/2# --># tan x = sqrt3/3# -->
#cos^2 x = 3/4 #

#(sqrt3/2)((sqrt3)/3) - 2(3/4) = 3/6 - 3/2 = -6/6 = - 1. OK#

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