How do you solve $cos(3x)=sqrt3/2$ and find all solutions in the interval $0<=x<360$ ?

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Answer 1 · 2017-04-06T05:15:34

$10^@; 110^@$

$cos 3x= sqrt3/2$
Trig table and unit circle give 2 solutions:
$3x = +- 30^@$

a. $3x = 30^@$ --> $x = 10^@$
b. $3x = - 30^@$ , or $3x = 330^@$ (co-terminal) -->
$x = 110^@$

How do you solve cos(3x)=sqrt3/2cos(3x)=sqrt3/2 and find all solutions in the interval 0<=x<3600<=x<360?