How do you solve #(tanx+1)^2=sec^2x-3#?

2 Answers
Gerardina C.
Oct 11, 2016

#x=arctan(-3/2)+kpi#

Explanation:

Since #color(red)(tanx=sinx/cosx) and color(green)(secx=1/cosx)#, you have:

#(color(red)(sinx/cosx)+1)^2=(color(green)(1/cosx))^2-3#

#((sinx+cosx)/cosx)^2=(1-3cos^2x)/cos^2x#

#(color(blue)(sin^2x+cos^2x)+2sinxcosx)/cos^2x=(1-3cos^2x)/cos^2x#

Since #color(blue)(sin^2x+cos^2x=1)#, you have:

#cancel(color(blue)1)+2sinxcosx=cancel(color(blue)1)-3cos^2x and cosx!=0#

#2sinxcosx+3cos^2x=0 and cosx!=0#

#cosx(2sinx+3cosx)=0 and cosx!=0#

#2sinx+3cosx=0#

then you can divide by cos x to obtain:

#2color(red)(sinx/cosx)+3cosx/cosx=0#

#2color(red)tanx=-3#

#tanx=-3/2#

#x=arctan(-3/2)+kpi#

Nghi N.
Oct 11, 2016

# -56^@31 + k180^@#

Explanation:

Use trig identity: #(tan^2 x + 1) = sec^2 x#
#(tan x + 1)^2 = tan^2 x + 1 + 2tan x = sec^2 x - 3 #
#sec^2 x + 2tan x = sec^2 x - 3#
Cancel #sec^2 x# from both sides, we get
2tan x = -3
tan x = - 1.5
Calculator gives
#x = - 56^@31#.
General answers:
#x = - 56^@31 + k180^@#
Check by calculator.
x = - 56.31 --> tan x = -1. 5 --> (tan x + 1)^2 = 0.25
sec^2 x = 1/(cos^2 x) = 1/0.30 = 3.25.
sec^2 x - 3 = 3.25 - 3 = 0.25. OK

Related questions
See all questions in Solving Trigonometric Equations
Impact of this question
1653 views around the world
You can reuse this answer
Creative Commons License