Posted by **-** on Sunday, November 15, 2009 at 12:40am.

Solve 3sin2x - 1 = 0

Interval: [0, 2 pi]

Solve 6cos^2x + 5cosx - 6 = 0

- math -
**drwls**, Sunday, November 15, 2009 at 1:35am
For the first one, sin2x = 1/3

2x = 0.33984, 2.80176 or 6.62302 radians (and one more less than 4 pi)

x = 0.16992, 1.4088, or 3.3115

(and one more less than 2 pi that I will leave you to figure out)

For the second question, factor into

(2cosx +3)(3cosx -2) = 0

Then set each factor = 0 and solve for cos x.

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