SOLVE to the nearest four decimals?


sqrt(3)*(cotx)^2+cotx=13(cotx)2+cotx=1

2 Answers
May 17, 2018

x = 62^@3393 + k180^@x=623393+k180
3x = - 42^@2363 + k180^@3x=422363+k180

Explanation:

Method 1
sqrt3(cos^2 x/(sin^2 x)) + cos x/(sin x) = 13(cos2xsin2x)+cosxsinx=1
sqrt3.cos^2 x + sin x.cos x = sin^2 x3.cos2x+sinx.cosx=sin2x
Divide both side by cos x, (condition cos x != 0)
sqrt3 + tan x = tan^2 x3+tanx=tan2x
tan^2 x - tan x - sqrt3 = 0tan2xtanx3=0
Solve this quadratic equation for tan x.
D = d^2 = b^2 - 4ac = 1 + 4sqrt3 = 7.9282 --> d = +- 2.8157d=±2.8157
There are 2 real roots:
tan x = -b/(2a) +- d/(2a) = 1/2 +- 2.8157/2tanx=b2a±d2a=12±2.81572
tan x = 0.5 + 1.4079 = 1.9079tanx=0.5+1.4079=1.9079
tan x = 0.5 - 1.4079 = - 0.9079tanx=0.51.4079=0.9079
a. tan x = 1.9079tanx=1.9079
Calculator and unit circle give:
x = 62^@3393 + k180^@x=623393+k180
b. tan x = - 0.9079tanx=0.9079
x = -42^@2363 + k180^@x=422363+k180

May 17, 2018

x = 62^@3405 + k180^@x=623405+k180
x = -42^@2614 + k180^@x=422614+k180

Explanation:

Method 2
Call cot x = t, we get a quadratic equation to solve:
sqrt3t^2 + t - 1 = 03t2+t1=0
D = d^2 = b^2 - 4ac = 1 + 4sqrt3 = 7.9282D=d2=b24ac=1+43=7.9282 --> d = +- 2.8157d=±2.8157
There are 2 real roots:
t = -b/(2a) +- d/(2a) = - 1/(2sqrt3) +- 2.8157/(2sqrt3) =t=b2a±d2a=123±2.815723=
= - 0.2887 +- 0.8128=0.2887±0.8128
cot x = t = 0.5241 --> tan x = 1.9080tanx=1.9080
cot x = t = - 1.1015 --> tan x = - 0.9078tanx=0.9078
a. tan x = 1.9080
x = 62^@3405 + k180^@x=623405+k180
b. tan x = - 0.9087
3x = - 42^@2614 + k180^@3x=422614+k180