# How you solve system of eqs  tgx+tgy=1+root3 (tgx)^2+(tgy)^2=4

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You may use the following substitutions, such that:

`tan x = a, tan y = b`

Replacing a for tan x and b for tan y yields:

`{(a + b = 1 + sqrt3),(a^2 + b^2 = 4):}`

You should use the following formula, such that:

`a^2 + b^2 = (a + b)^2 - 2ab`

Replacing `(a + b)^2 - 2ab` for `a^2 + b^2` yields:

`{(a + b = 1 + sqrt3),((a + b)^2 - 2ab = 4):}`

Replacing `1 + sqrt3` for `a + b` in the bottom equation yields:

`{(a + b = 1 + sqrt3),((1 + sqrt3)^2 - 2ab = 4):}`

`{(a + b = 1 + sqrt3),(2ab = (1 + sqrt3)^2 - 4):}`

`{(a + b = 1 + sqrt3),(2ab = 1 + 2sqrt3 + 3 - 4):}`

Reducing duplicate members yields:

`{(a + b = 1 + sqrt3),(2ab = 2sqrt3):}`

`{(a + b = 1 + sqrt3),(ab = sqrt3):}`

Using Lagrange resolvents yields:

`p = a + b = -(1 + sqrt3), q = ab = sqrt3`

`x^2 + px + q = 0 => x^2 - (1 + sqrt3)x + sqrt3 = 0 x^2 - x - sqrt3*x + sqrt 3 = 0`

You need to group the terms such that:

`(x^2 - sqrt3*x) - (x - sqrt 3) = 0`

`x(x - sqrt 3) - (x - sqrt 3) = 0 => (x - sqrt 3)(x - 1) = 0`

Using the zero product rule, yields:

`{(x - sqrt 3 = 0),(x - 1 = 0):} => {(a = sqrt 3, b = 1),(a = 1, b = sqrt 3):}`

Replacing back tan x for a and tan y for b yields:

`{(tan x = sqrt 3, tan y = 1),(tan x = 1, tan y = sqrt 3):}`

`{(x = pi/3 + n*pi, y = pi/4 + n*pi),(x = pi/4 + n*pi, y = pi/3 + n*pi):}`

Hence, evaluating the solutions to the system of simultaneous equations yields `x = pi/3 + n*pi, x = pi/4 + n*pi, y = pi/4 + n*pi, y = pi/3 + n*pi.`

oldnick | (Level 1) Valedictorian

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`tanx+tany=1+sqrt(3)`

`tan^2x+tan^2y=4`

`(tanx+tany)^2= tan^2x+tan^2y + 2tanx tan y`

`(1+sqrt(3))^2=4+2tanx tany`

`4+2sqrt(3)=4+2tanx tany`  `rArr tanx tany=sqrt(3)`

`(1+sqrt(3))/(1-sqrt(3))=tan(x+y)` `=-(2+sqrt(3))`  `rArr x+y=105^0`

`tan y=tan(105^0-x)=` `(-(2+sqrt(3))-tan x)/(1-(2+sqrt(3))tanx)=`

`tanx tany=sqrt(3)` `rArr`  `-tanx (2+sqrt(3)+tanx)/(1-(2+sqrt(3))tanx)=sqrt(3)`

`-2tanx-sqrt(3)tanx-tan^2x=sqrt(3)-2sqrt(3)tanx-3tanx`

`tan^2x -(sqrt(3)+1)tanx+sqrt(3)=0`

`Delta=(sqrt(3)+1)^2-4sqrt(3)=4-2sqrt(3)=(1-sqrt(3))^2`

`tanx=(1+sqrt(3)+-(1-sqrt(3)))/2`

`tanx=1;tanx=sqrt(3)`

`symmetric..system....so:`

`x= pi/4+kpi; y=pi/3+kpi`

`or:`

`x=pi/3+kpi;y=pi/4+kpi`

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