How to solve this for 0 ≤ x ≤ 2π ?

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2 Answers
Apr 19, 2018

Given: #tan^2(x)+tan(x) = sqrt3tan(x) + sqrt3, 0 <= x < 2pi#

Add #-(sqrt3tan(x) + sqrt3)# to both sides:

#tan^2(x)+(1-sqrt3)tan(x) - sqrt3= 0, 0 <= x < 2pi#

Factor:

#(tan(x)+1)(tan(x) - sqrt3)= 0, 0 <= x < 2pi#

#tan(x) = -1# and #tan(x) = sqrt3, 0 <= x < 2pi#

#x = tan^-1(-1)# and #x = tan^-1(sqrt3), 0 <= x < 2pi#

#x = 3/4pi, 7/4pi, 1/3pi, and 4/3pi#

Apr 19, 2018

That's a quadratic equation in #\tan x.#

# tan ^2 x + \tan x = \sqrt{3} \tan x + \sqrt{3} #

# tan^2 x +(1-\sqrt{3}) \tanx - \sqrt{3} = 0#

Despite the square roots, this isn't too hard to factor:

# ( \tan x - \sqrt{3} )(\tan x + 1) = 0#

# \tan x = -1 # or # \tan x = \sqrt{3}#

My pet peeve is every problem in trig is 30,60,90 or 45,45,90; this one's both!

#x = frac{3\pi}{ 4}# or #x = frac{7 \pi}{4}# or #x=frac{pi}{4}# or #x=\frac{4\pi}{3}#