$a)3\cdot {\mathrm{tg}}^{2}x-\frac{1}{{\cos }^{2}x}=5$  

Założenia:

${\cos }^{2}x\ne 0$  

$\cos x\ne 0$  

$x\ne \frac{\pi }{2}+k\pi ,k\in C$  

 

$\frac{3{\sin }^{2}x}{{\cos }^{2}x}-\frac{1}{{\cos }^{2}x}=5$  

$\frac{3{\sin }^{2}x-1}{{\cos }^{2}x}=5$  

$\frac{3{\sin }^{2}x-1}{1-{\sin }^{2}x}=5$  

Podstawienie pomocnicze:

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

 

$\frac{3t-1}{1-t}=5$  

$3t-1=5\left(1-t\right)$  

$3t-1=5-5t$  

$8t=6$  

$t=\frac{6}{8}$  

$t=\frac{3}{4}$  

Stąd:

${\sin }^{2}x=\frac{3}{4}$  

$\left|\sin x\right|=\frac{\sqrt{3}}{2}$  

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

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

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

$b)\sin x+{\cos }^{2}x=\frac{1}{4}$  

$\sin x+1-{\sin }^{2}x=\frac{1}{4}$  

Podstawie pomocnicze:

$t=\sin x,t\in \left[-1,1\right]$  

 

$t+1-t^2=\frac{1}{4}\mid \cdot 4$  

$4t+4-4t^2=1$  

$4t^2-4t-3=0$  

$\Delta ={\left(-4\right)}^2-4\cdot 4\cdot \left(-3\right)=16+48=64$  

$\sqrt{\Delta }=\sqrt{64}=8$  

$t_1=\frac{4-8}{8}\vee t_2=\frac{4+8}{8}>1$   

$t_1=-\frac{4}{8}$  

$\sin x=-\frac{1}{2}$  

$x_1=-\frac{\pi }{6}+2k\pi \vee x_2=\pi +\frac{\pi }{6}+2k\pi$  

$x_1=-\frac{\pi }{6}+2k\pi \vee x_2=\frac{7\pi }{6}+2k\pi ,k\in C$  

 

$c)2\mathrm{tg}x+3\mathrm{ctg}x+5=0$  

$2\mathrm{tg}x+\frac{3}{\mathrm{tg}x}+5=0$  

Podstawienie pomocnicze:

$t=\mathrm{tg}x,t\ne 0$   

 

$2t+\frac{3}{t}+5=0\mid \cdot t$  

$2t^2+3+5t=0$  

$2t^2+2t+3t+3=0$  

$2t\left(t+1\right)+3\left(t+1\right)=0$  

$\left(t+1\right)\left(2t+3\right)=0$  

$t_1=-1\vee t_2=-\frac{3}{2}$  

$\mathrm{tg}x=-1\vee \mathrm{tg}x=-\frac{3}{2}$  

$x_1=-\frac{\pi }{4}+k\pi \vee x_2=\alpha +k\pi ,\mathrm{tg}\alpha =-\frac{3}{2},k\in C$  

 

$d){\mathrm{ctg}}^{2}x-3{\mathrm{tg}}^{2}x=2$  

${\mathrm{ctg}}^{2}x-\frac{3}{{\mathrm{ctg}}^{2}x}=2$   

Podstawienie pomocnicze:

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

 

$t-\frac{3}{t}=2\mid \cdot t$   

$t^2-3=2t$  

$t^2-2t-3=0$  

$t^2+t-3t-3=0$  

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

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

$t_1=-1\vee t_2=3$

Bierzemy pod uwagę tylko dodatnie:

${\mathrm{ctg}}^{2}x=3$   

$\left|\mathrm{ctg}x\right|=\sqrt{3}$  

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

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