How do you solve $2cos(x/3)+sqrt2=0$ ?

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Answer 1 · 2016-09-20T04:59:05

$x = (9pi)/4 + 6kpi$
$x = (15pi)/4 + 6kpi$

$2cos (x/3) + sqrt2 = 0$
$cos (x/3) = -sqrt2/2$
Trig Table and unit circle -->
$x/3 = +- (3pi)/4$

a. $x/3 = (3pi)/4 + 2kpi$
$x = 9pi/4 + 6kpi$
b. $x/3 = (5pi)/4 + 2kpi$
(Arc $(-3pi)/4$ is co-terminal to arc $(5pi)/4$ )
$x = 15pi/4 + 6kpi$ ,

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