Question

1) 4 cot²x = 3 cosec²x

Answer 1

$1 + cot^2(x) = cosec^2(x) \\ \\ 4 cot^2(x) = 3 cosec^2(x) \\ \\ 4cot^2(x) = 3(1 + cot^2(x)) \\ \\ 4cot^2(x) = 3 + 3cot^2(x) \\ \\ cot^2(x) = 3 \\ \\ cot(x) = (-\sqrt{3}, \sqrt{3})$

So,
$x = (n \pi + \frac{ \pi }{6}, n \pi - \frac{ \pi }{6})$

Answer 2

[tex]\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[/tex]

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