Question 927955
Find values of x between 0 degrees and 360 degrees.
(3cos(x)-2)(2cos(x)+1)=0
6cos^2(x)-cosx-2=0
(3cos(x)-2)(2cos(x)+1)=0
..
3cos(x)-2=0
cos(x)=2/3
x≈49.19˚,311.81˚(in quadrants I and IV where cos>0)
or
2cos(x)+1=0
cos(x)=-1/2
x=120&#730;, 240&#730; (in quadrants II and III where cos<0)