Solve for x the equation tan(x+pi/3)=tan(pi/2-x), if 0<x<pi?
We have to solve tan(x+pi/3)=tan(pi/2-x) for values of x that satisfy 0<x<pi.
In the specified range, each value of x has a unique value of tan x.
This allows us to equate x + pi/3 and pi/2 - x to arrive at x.
x + pi/3 = pi/2 - x
=> 2x = pi/2 - pi/3
=> 2x = (3pi - 2pi)/6
=> x = pi/12
The required value of x = pi/12
The equation could be solved in 2 ways, at least.
(x + pi/3) = arctan[tan(pi/2-x)] + kpi
But tan (arctan x) = x
(x + pi/3) = (pi/2-x) + kpi
We'll add x both sides:
2x = pi/2 - pi/3 + kpi
2x = pi/6 + kpi
x = pi/12 + kpi
Since x has to belong to the opened interval (0,pi), then x = pi/12.