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

How do you solve 10tanx-5=010tanx-5=0?