Question 723889
Find all values of &#952;, to the nearest degree, that satisfy the equation 7cos cubed&#952; = 5cos squared&#952; +cos&#952; in the interval 0< &#952; < 360
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7cos^3x=5cos^2x+cosx
7cos^3x-5cos^2x-cosx=0
cosx(7cos^2x-5cosx-1)=0
cosx=0
x=90º,270º
..
7cos^2x-5cosx-1)=0
solve for cos x by quadratic formula:
cosx&#8776;-0.1629
using inverse cos key on calculator:
x&#8776;99º, 261º( in quadrants II and III where cos<0)
or
cosx&#8776;0.8772
using inverse cos key on calculator:
x&#8776;28.7º, 331º( in quadrants I and IV where cos>0)
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all values of &#952; in the interval 0< &#952; < 360: 90º, 270º, 99º, 261º, 28.7º, 331º