SOLUTION: I need help figuring out this trigonometry problem.
Fin,d to the nearest degree, all values of x in the interval 0 degrees < x < 360 degrees that satisfy the equation 6cos^2x - 5s
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Question 436688: I need help figuring out this trigonometry problem.
Fin,d to the nearest degree, all values of x in the interval 0 degrees < x < 360 degrees that satisfy the equation 6cos^2x - 5sinx - 5 = 0
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
I need help figuring out this trigonometry problem.
Fin,d to the nearest degree, all values of x in the interval 0 degrees < x < 360 degrees that satisfy the equation 6cos^2x - 5sinx - 5 = 0
===================
6(1-sin^2)-5sin - 5 = 0
-6sin^2-5sin+1 = 0
----
6sin^2+5sin-1 = 0
(6sin-1)(sin+1) = 0
sin(x) = 1/6 or sin(x) = -1
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x = 9.59 degrees or 170.41 degrees or 270 degrees.
====================================================
Cheers,
Stan H.
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