Question
Solve the trig equation exactly in the interval 0≤x≤2π.
tan x + tan π/3
------------------ = square root 3
1 - tan x tan π/3
tan x + tan π/3
------------------ = square root 3
1 - tan x tan π/3
Answers
Reiny
you should recognize that they are using the formula
tan(A+B) = (tanA + tanB)/(1 - tanAtanB)
so
(tanx + tan π/3)/(1 - tanxtan(π/3) = √3
becomes
tan(x+π/3) = √3 ,
I know tan 60° or tan 240° = √3 OR tan π/3 or tan 4π/3 = √3
so x+π/3 = π/3 OR x+π/3 = 4π/3
x = 0 or x = π
but the period of tanx is π
so
x = 0 , π , 2π
tan(A+B) = (tanA + tanB)/(1 - tanAtanB)
so
(tanx + tan π/3)/(1 - tanxtan(π/3) = √3
becomes
tan(x+π/3) = √3 ,
I know tan 60° or tan 240° = √3 OR tan π/3 or tan 4π/3 = √3
so x+π/3 = π/3 OR x+π/3 = 4π/3
x = 0 or x = π
but the period of tanx is π
so
x = 0 , π , 2π