Question

SOLVE to the nearest four decimals?

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

Answer 1

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

Explanation:

Method 1
`sqrt3(cos^2 x/(sin^2 x)) + cos x/(sin x) = 1`
`sqrt3.cos^2 x + sin x.cos x = sin^2 x`
Divide both side by cos x, (condition cos x != 0)
`sqrt3 + tan x = tan^2 x`
`tan^2 x - tan x - sqrt3 = 0`
Solve this quadratic equation for tan x.
D = d^2 = b^2 - 4ac = 1 + 4sqrt3 = 7.9282 --> `d = +- 2.8157`
There are 2 real roots:
`tan x = -b/(2a) +- d/(2a) = 1/2 +- 2.8157/2`
`tan x = 0.5 + 1.4079 = 1.9079`
`tan x = 0.5 - 1.4079 = - 0.9079`
a. `tan x = 1.9079`
Calculator and unit circle give:
`x = 62^@3393 + k180^@`
b. `tan x = - 0.9079`
`x = -42^@2363 + k180^@`

Answer 2

`x = 62^@3405 + k180^@`
`x = -42^@2614 + k180^@`

Explanation:

Method 2
Call cot x = t, we get a quadratic equation to solve:
`sqrt3t^2 + t - 1 = 0`
`D = d^2 = b^2 - 4ac = 1 + 4sqrt3 = 7.9282` --> `d = +- 2.8157`
There are 2 real roots:
`t = -b/(2a) +- d/(2a) = - 1/(2sqrt3) +- 2.8157/(2sqrt3) =`
`= - 0.2887 +- 0.8128`
cot x = t = 0.5241 --> `tan x = 1.9080`
cot x = t = - 1.1015 --> `tan x = - 0.9078`
a. tan x = 1.9080
`x = 62^@3405 + k180^@`
b. tan x = - 0.9087
`3x = - 42^@2614 + k180^@`