$a){\mathrm{ctg}}^{4}x-2{\mathrm{ctg}}^{2}x-3=0$  

Podstawienie pomocnicze:

$t={\mathrm{ctg}}^{2}x,t\in \left[0,\infty \right)$   

$t^2-2t-3=0$  

$t^2+t-3t-3=0$  

$t\left(t+1\right)-3\left(t+1\right)=0$  

$\left(t+1\right)\left(t-3\right)=0$  

$t_1=-1\vee t_2=3$   

Bierzemy pod uwagę tylko dodatnie t:

$t=3$   

${\mathrm{ctg}}^{2}x=3$  

${\mathrm{ctg}}^{2}x-3=0$  

$\left(\mathrm{ctg}x-\sqrt{3}\right)\left(\mathrm{ctg}x+\sqrt{3}\right)=0$  

$\mathrm{ctg}x=\sqrt{3}\vee \mathrm{ctg}x=-\sqrt{3}$  

$x_1=\frac{\pi }{6}+k\pi \vee x_2=-\frac{\pi }{6}+k\pi$  

 

$b)4{\sin }^{4}x-5{\cos }^{2}x-1=0$  

$4{\sin }^{4}x-5\left(1-{\sin }^{2}x\right)-1=0$  

$4{\sin }^{4}x-5+5{\sin }^{2}x-1=0$  

$4{\sin }^{4}x+5{\sin }^{2}x-6=0$  

Podstawienie pomocnicze:

$t={\sin }^{2}x,t\in \left[0,1\right]$  

 

$4t^2+5t-6=0$  

$4t^2+8t-3t-6=0$   

$4t\left(t+2\right)-3\left(t+2\right)=0$  

$\left(t+2\right)\left(4t-3\right)=0$  

$t+2=0\vee 4t-3=0$  

$t=-2\vee t=\frac{3}{4}$  

Uwzględniając dziedzinę otrzymujemy, że:

$t=\frac{3}{4}$  

${\sin }^{2}x=\frac{3}{4}$  

${\sin }^{2}x-\frac{3}{4}=0$  

$\left(\sin x-\frac{\sqrt{3}}{2}\right)\left(\sin x+\frac{\sqrt{3}}{2}\right)=0$  

$\sin x=\frac{\sqrt{3}}{2}\vee \sin x=-\frac{\sqrt{3}}{2}$  

$x_1=\frac{\pi }{3}+2k\pi \vee x_2=\pi -\frac{\pi }{3}+2k\pi \vee x_3=-\frac{\pi }{3}+2k\pi \vee x_4=-\pi +\frac{\pi }{3}+2k\pi$  

$x_1=\frac{\pi }{3}+2k\pi \vee x_2=\frac{2\pi }{3}+2k\pi \vee x_3=-\frac{\pi }{3}+2k\pi \vee x_4=-\frac{2\pi }{3}+2k\pi ,k\in C$