How do you solve $3 tan x + 1 = 13$ $3tanx+1=13$ ?

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Answer 1 · 2018-05-03T17:55:41

$x ~~ 1.3258... + npi$ , $n in ZZ$ .

We have

$3tanx+1=13$
$3tanx=12$
$tanx=4$

Now, there is no 'nice' form of the answer. Instead, we can just accept that:

$x=arctan4$

And since the tangent function is periodic with period $rho=npi$ for an integer $n$ , the answer we seek is

$x=arctan4+npi ~~ 1.3258...+npi$ , $n in ZZ$ .

Answer 2 · 2018-05-04T03:44:10

$x = 75^@96 + k180^@$

$3tan x + 1 = 13$

$3tan x = 12$

$tan x = 4$

Calculator and unit circle give:

$x = 75.96^@ + k180^@, k in ZZ$

How do you solve 3tanx+1=133tanx+1=133tanx+1=13?