Solve The Following Expression.

$\mathrm{sin3x}+\mathrm{cos2x}=0$$\mathrm{sin3x}=-\mathrm{cos2x}$$3\mathrm{sinx}-4{\sin }^{3}x = -(1-2{\sin }^{2}x)$$3\mathrm{sinx}-4{\sin }^{3}x = -1+2{\sin }^{2}x$$4{\sin }^{3}x+2{\sin }^{2}x-3\mathrm{sinx}-1=0$$\mathrm{This} \mathrm{is} a \mathrm{cubic} \mathrm{equation}. \mathrm{let} a =\mathrm{sinx}$$4a^3+2a^2-3a-1=0$$\mathrm{possible} \mathrm{roots} \mathrm{are} -1 \mathrm{or} 1$$a = -1 \mathrm{satisfies} \mathrm{the} \mathrm{equation}$$(a+1)(4a^2-2a-1)=0$$\mathrm{so} a=-1 \mathrm{and} \mathrm{on} \mathrm{solving} \mathrm{the} \mathrm{quadratic},$$a = \frac{2+\sqrt{20}}{8} \mathrm{or} \frac{2-\sqrt{20}}{8} =\frac{1+\sqrt{5}}{4} \mathrm{or} a = \frac{1-\sqrt{5}}{4}$$\mathrm{But} a=\mathrm{sinx} , \mathrm{so} \mathrm{sinx} = -1,\frac{1+\sqrt{5}}{4},\frac{1-\sqrt{5}}{4}$">