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hellohi hellohi
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13 years ago Edited: 7 years ago, duddy
6cos^2theta+sin theta -4= 0 solve for theta in radians, 0 <= 0 <= 2pi.

Thanks guys.
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13 years ago
Quote from: hellohi (13 years ago)
6cos^2theta+sin theta -4= 0 solve for theta in radians, 0 <= 0 <= 2pi.

Thanks guys.

6cos2x + sin x - 4 = 0
6(1-sin2x) + sin x - 4 = 0
6sin2x - sin x - 2 = 0
6sin2x - 4sin x + 3 sin x -2 = 0
2sin x (3 sin x - 2)+1(3 sin x -2)=0
(2sinx+1)(3sinx-2)=0
gives sin x = -1/2 or 2/3

we will take both values because the value of x lies between 0 and 2pi

once you get the values of angles in degrees, you can convert it in radians using the formula

degree = radians * (180/pi)
I don't feel like riding until everything blurs.

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