How do you solve #6tan^2x=5tanx+sec^2x#?

1 Answer
Nghi N.
Sep 11, 2016

#49^@48 + k180#
#-9^@65 + k180#

Explanation:

#6tan^2x = 5tan x + sec^2 x#
Call tan x = t and use the trig identity:
#sec^2 t = 1 + tan^2 t#
#6t^2 = 5t + 1 + t^2#
Solve this quadratic equation for t by the improved quadratic formula (Socratic Search)
#5t^2 - 5t - 1 = 0#
#D = d^2 = b^2 - 4ac = 25 + 20 = 45# --> #d = +- 3sqrt5#
There are 2 real roots:
#t = -b/(2a) +- d/(2a) =#
#t = 5/10 +- (3sqrt5)/10 = 1/2 +- (3sqrt5)/10#
t1 = 0.5 + 0.67 = 1.17, and t2 = 0.5 - 0.67 = - 0.17.
Calculator -->
a tan x1 = 1.17 --> #x1 = 49^@48 + k180#
b. tan x2 = -0.17 --> #x2 = -9^@65 + k180#

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