SOLUTION: Approximate to the nearest degree, the solutions of the equation in the interval [0�, 360�). 6cos^2(x) = 11sin(x)+4

Question 524724: Approximate to the nearest degree, the solutions of the equation in the interval [0�, 360�).
6cos^2(x) = 11sin(x)+4
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Approximate to the nearest degree, the solutions of the equation in the interval [0�, 360�).
6cos^2(x) = 11sin(x)+4
**
6(1-sin^2(x) = 11sin(x)+4
6(1-sin^2(x) = 11sin(x)+4
6-6sin^2(x)= 11sin(x)+4
6sin^2(x)+11sin(x)-2=0
(sin(x)+2)(6sin(x)-1)=0
..
sin(x)+2=0
sin(x)=-2 (reject [-1≤sinx≤1]
..
6sin(x)-1=0
sinx=1/6
x≈10� or 170�
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