# Solve for x the equation tan(x+pi/3)=tan(pi/2-x), if 0<x<pi?

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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.**