$\underline{k\in \mathbb{C}}$  

 

$a)$   

$2{\cos }^{2}x-\cos x=1$  

$t=\cos x$  

$2t^2-t-1=0$  

$\Delta =1+8=9$  

$\sqrt{\Delta }=3$  

$t_1=\frac{1-3}{4}=-\frac{1}{2}$  

$t_2=\frac{1+3}{4}=1$  

wracając do podstawienia mamy

$\cos x=-\frac{1}{2}=-\cos \left(\pi +x\right)\vee \cos x=1$  

$\cos \left(\pi +x\right)=\frac{1}{2}\vee \cos x=1$  

$\pi +x=\frac{\pi }{3}+2k\pi \vee \pi +x=2\pi -\frac{\pi }{3}+2k\pi \vee x=2k\pi$  

$x=-\frac{2}{3}\pi +2k\pi \vee x=-\frac{4}{3}\pi +2k\pi \vee x=2k\pi$  

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$b)$  

$2{\cos }^{2}x+5\cos x+2=0$  

$t=\cos x$  

$2t^2+5t+2=0$  

$\Delta =25-16=9$  

$\sqrt{\Delta }=3$  

$t_1=\frac{-5-3}{4}=-2$  

$t_2=\frac{-5+3}{4}=-\frac{1}{2}$  

wracając do podstawienia mamy

$\cos x=-2\text{(sprzecznosˊcˊ)}\vee \cos x=-\frac{1}{2}$  

$\cos x=-\frac{1}{2}=-\cos \left(\pi +x\right)$  

$\cos \left(\pi +x\right)=\frac{1}{2}$  

$\pi +x=\frac{\pi }{3}+2k\pi \vee \pi +x=2\pi -\frac{\pi }{3}+2k\pi$  

$x=-\frac{2}{3}\pi +2k\pi \vee x=-\frac{4}{3}\pi +2k\pi$  

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$c)$  

${\mathrm{tg}}^{2}x+2\mathrm{tg}x+1=0$  

 

$D:$  

$x\ne \frac{\pi }{2}+k\pi$  

 

$t=\mathrm{tg}x$  

$t^2+2t+1=0$  

$\Delta =4-4=0$  

$t=-\frac{2}{2}=-1$  

wracając do podstawienia mamy

$\mathrm{tg}x=-1=-\mathrm{tg}\left(-x\right)$   

$-x=\frac{\pi }{4}+k\pi$  

$x=-\frac{\pi }{4}-k\pi$  

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$d)$  

$\frac{\cos x+{\cos }^{2}x+\frac{1}{4}}{\sin x}=0$  

 

$D:$  

$\sin x\ne 0$  

$x\ne k\pi$  

 

$t=\cos x$  

$t^2+t+\frac{1}{4}=0$  

$\Delta =1-1=0$  

$t=-\frac{1}{2}$  

$\cos x=-\frac{1}{2}=-\cos \left(\pi +x\right)$   

$\cos \left(\pi +x\right)=\frac{1}{2}$  

$\pi +x=\frac{\pi }{3}+2k\pi \vee \pi +x=2\pi -\frac{\pi }{3}+2k\pi$  

$x=-\frac{2}{3}\pi +2k\pi \vee x=-\frac{4}{3}\pi +2k\pi$