In the range \(0\leq x<2\pi\), the equation \[\sin^2x+3\sin x\cos x+2\cos^2 x=0\] has

\(1\) solution,

\(2\) solutions,

\(3\) solutions,

\(4\) solutions.

If \(\cos x = 0, \sin x \neq 0\), so the equation does not hold. Thus we are free to divide by \(\cos^2x\), since it must be non-zero, giving \[\tan^2x + 3\tan x + 2=0,\]

which factorises into \[(\tan x +1)(\tan x +2 ) = 0.\]

So our solutions will be those for \(\tan x = -1, \tan x = -2.\)

Since \(y=\tan x\) has a period of \(\pi\), there will therefore be four solutions in total, which means the answer is (d).