Posted by bkue on .
solve for the equation x
sin x= .35
180<x<270
sin x= .2
90<x<180
cos x = .72
180< x<270
please explain, thanks :)

trig 
Reiny,
Here is a standard way to solve these kind of problems, I will do the first one
sinx = .35 , 180 < x <270
1. Find the acute angle, sometimes called the "angle in standard position"
by using the positive decimal
if sinx = .35, x = 20.5°
but x is supposed to be in III , then
x = 180 + 20.5 = 200.5°
check: on calculator, sin 200.5 = .3502..
try the others the same way, check the answer with your calculator.