Question

tan^2x + (1-√3)tanx -√3=0
solve for x

Answer 1

Given,

⇒tan²x + ( 1 - √3 ) tan x - √3 = 0


⇒ tan²x + tan x - √3 tan x - √3 = 0

Taking out tan x as common from first two terms and ( -√3 ) from last two terms,

⇒ tan x ( tan x + 1 ) - √3 ( tan x + 1 ) = 0

Taking out ( tan x + 1 ) as common,

⇒ ( tan x + 1 ) ( tan x - √3 ) = 0

⇒ ( tan x + 1 ) = 0 ÷ ( tan x - √3 )

⇒ ( tan x + 1 ) = 0

•°• tan x = -1

" Or "

⇒ ( tan x + 1 ) ( tan x - √3 ) = 0

⇒ ( tan x - √3 ) = 0 ÷ ( tan x + 1 )

⇒ ( tan x - √3 ) = 0

•°• tan x = √3.

Hence, tan x = -1 or √3.

When , tan x = -1

We know that , tan 135° = -1

So, x = -135°

When , tan x = √3

We know that,

⇒ tan 60° = √3

Or

⇒tan -120° = √3

•°• x = 60° or -120°

Hence, x = 60° , 135° , -120°

Answer 2

Answer:

Step-by-step explanation:

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

from above equation, we get

tanx =  $\frac{ -(1 -\sqrt{3}) \frac{+}{} \sqrt{(1-\sqrt{3})^2 - 4(1)(-\sqrt{3}) } }{2(1)}$

tanx = -1 , $\sqrt{3}$

x = n$\pi$ $-\frac{\pi}{4}$ or n$\pi$ + $\frac{\pi }{3}$

where n ∈ Z