Answers

2014-08-23T20:00:31+05:30
1 + cot^2(x) = cosec^2(x) \\  \\ 4 cot^2(x) = 3 cosec^2(x) \\  \\ 4cot^2(x) = 3(1 + cot^2(x)) \\  \\ 4cot^2(x) = 3 + 3cot^2(x) \\  \\ cot^2(x) = 3 \\  \\ cot(x) =  (-\sqrt{3}, \sqrt{3})

So,
x = (n \pi +  \frac{ \pi }{6}, n \pi -  \frac{ \pi }{6})
3 3 3
This should be pi +- not 2pi +- because, tan(x) = y has a general solution x = arctan(y) + n pi
2014-08-24T03:09:50+05:30

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\frac{4 cos^{2} x }{sin^{2} x} = \frac{3}{sin^{2} x} \\ \\ 
cos^{2} x = \frac{3}{4} \\ \\ cos\ x = +-\frac{\sqrt{3}}{2} \\ \\ x = 2\Pi +- 30\ deg\\ \\ and\ x\ =\ \Pi\ +-\ 30\ deg

So there are 4 possible solutions in [0, 2π]

2 5 2
there are three solutions not four..... in [0, 2pi] coz 2pi+30 is not in [0, 2pi]
30 deg, 150 deg, 210 deg, 330 deg satisfy the given problem. chek urself. if u want, to verify
i am not taking about this thing, i referred the lines you wrote, x = 2pi +- 30deg, then x=pi +- 30deg, and then u wrote 4 solution so anyone will simply think, that u r talking about these four solutions u listed here