Find all solutions in the interval ​[0,2π​). tangent squared x equals 1

Question

I do not understand how to solve this equation and I need help.

Answer 1

solve tan²x=1

tan²x-1=0

factor

(tanx+1)(tanx-1)=0

tanx+1=0

tanx=-1

tan=sin/cos

tan is -1 at 3∏/4 and at 7∏/4 or 135º and 315º

tanx-1=0

tanx=1

tan is 1 at ∏/4 and at 5∏/4 or 45º and 225º

also, the answers start at ∏/4 and increase by 2∏/4

Answer 2

You have to solve equation

tan²(x)=1.

This equation has many (basically infinite) number of solutions. You have to find onlys those solutions, which belong to interval [0, 2π]

If you take square root from both sides, you will have  two equations:

a) tan(x)=1  b) tan(x)=-1

x=arctan(1)+kπ, k=0, ±1, ±2,..  (tan x isπ is a periodic fucntion and π is its period)

x=π/4+kπ, k=0, ±1, ±2

If k=0, x=π/4∈[0, 2π], 

if  k=1, x=π/4+π=5π/4 ∈ [0, 2π]

 

if  k=2, x=π/4+2π=9π/4 ∉ [0, 2π]

So we have two solutions x=π/4 and x=5π/4 from interval [0, 2π]

Now, in similar way, solve equation

b) tan(x)=-1