Question 594974
Solve algebraically for all values of x in the interval 0&#8804; x < 360.2 sin2 x&#8722; 4 sin x = cos2 x &#8722; 2
Express your answers to the nearest degree.
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2 sin2 x&#8722; 4 sin x = cos2 x &#8722; 2
2sin^2-4sin=cos^2-2
2sin^2-4sin=1-sin^2-2
3sin^2-4sin+1=0
(3sinx-1)(sinx-1)=0
..
3sinx+1=0
sinx=-1/3
x&#8776;199º and 341º (in quadrants III and IV where sin<0)
..
sinx-1=0
sinx=1
x=90º
ans:
x=90º, 199º and 341º