Question 807942
Find the values of theta in [0 degrees, 360degrees) to the nearest tenth that satisfies the following equation
6 sin^2theta +sin theta=2
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use x for theta
{{{6sin^2(x)+sin(x)-2=0}}}
(3sin(x)+2)(2sin(x)-1)=0
..
3sinx+2=0
sinx=-2/3
x=221.81&#730;, 318.19&#730; (In quadrants III and IV where sin<0)
or
2sin(x)-1=0
sinx=1/2
x=30&#730;, 150&#730; (In quadrants I and II where sin>0)