How do you solve #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#