# What is the solution of equation tan^2x-18tanx+72=0?

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### 2 Answers

We have to solve (tan x)^2 - 18*tan x + 72 = 0

Let tan x = y

=> y^2 - 18y + 72 = 0

=> y^2 - 12y - 6y + 72 = 0

=> y(y - 12) - 6(y - 12) = 0

=> (y - 6)(y - 12) = 0

=> y = 6 and y = 12

As y = tan x

tan x = 6 and tan x = 12

x = arc tan 6 = 80.53 and x = arc tan 12 = 85.23

**As the tan function is periodic we get x = 80.53 + n*180 degrees and x = 85.23 + n*180 degrees.**

To solve the equation, we'll replace tan x by t:

The equivalent equation is:

t^2 - 18t + 72 = 0

We'll apply quadratic formula to find the roots:

t1 = [18+sqrt(324 - 288)]/2

t1 = (18+6)/2

t1 = 24/2

t1 = 12

t2 = (18-6)/2

t2 = 6

But tan x = t1 => tan x = 12 => x1 = arctan 12 + k*pi

tan x = t2 => tan x = 6 => x2 = arctan 6 + k*pi

**The solutions of the equation belong to the reunion of sets: {arctan 6 + k*pi}U{arctan 12 + k*pi}.**