How do you solve #10tanx-5=0#?

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Answer 1 · 2016-11-11T16:33:16

#26^@57 + k180^@#

10tan x - 5 = 0
#tan x = 5/10 = 1/2 = 0.50#
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#tan x = 0.50 -->arc # x = 26^@57 + k180^@#

Answer 2 · 2016-11-11T16:38:00

#(26.56)^o and (206.56)^o#

10 tan x -5=0 #->#tan x= #5/10= 1/2#

x= arctan#1/2# = #(26.56)^o and (180+26.56)^o or (206.56)^o#

These two are the principal solution of the equation