Можете пожалуйста помочь все что сможете?

Bb $2b).\quad 3sin^2x-cosx+1=0\\\\3(1-cos^2x)-cosx+1=0\\\\t=cosx,\; 3t^2+t-4=0,t_1=1,t_2=-\frac{4}{3}\\\\cosx=1,\; x=2\pi n,n\in Z\\\\cosx=-\frac{4}{3}\; net\; reshenij(|cosx| \leq 1)\\\\3b).\quad 3sin^2x+2\sqrt3sinxcosx+cos^2x=0\; |:cos^2x\ne 0\\\\3tg^2x+2\sqrt3tgx+1=0\\\\t=tgx,\; 3t^2+2\sqrt3t+1=0$

$D/4=3-3=0,\; (\sqrt3t+1)^2=0,\; t=-\frac{1}{\sqrt3}\\\\tgx=-\frac{1}{\sqrt3},\; x=arctg(-\frac{1}{\sqrt3})+\pi n=-\frac{\pi}{6}+\pi n,n\in Z\\\\4).\quad a)\; sinx=-0,5,\; x=(-1)^{n}arcsin(-0,5)+\pi n=\\\\=(-1)^{n}(-\frac{\pi}{6})+\pi n=(-1)^{n+1}\frac{\pi}{6}+\pi n,n\in Z\\\\b)cosx=\frac{1}{3},x=\pm arccos\frac{1}{3}+2\pi n,n\in Z\\\\c)tgx=-3,\; x=arctg(-3)+\pi n=-arctg3+\pi n,n\in Z\\\\5).\quad sinx+cosx=1\\\\sinx+sin(\frac{\pi}{2}-x)=1$

$2sin\frac{\pi}{4}\cdot cos(x-\frac{\pi}{4})=1\\\\2\cdot \frac{\sqrt2}{2}cos(x-\frac{\pi}{4})=1\\\\cos(x-\frac{\pi}{4})=\frac{1}{\sqrt2}\\\\x-\frac{\pi}{4}=\pm \frac{\pi}{4}+2\pi n\\\\x=\frac{\pi}{4}\pm \frac{\pi}{4}+2\pi n= \left \{ {{\frac{\pi}{2}+2\pi n,n\in Z} \atop {2\pi m,m\in Z}} \right.$