Miałam do rozwiązania jedną równość i trzy nierówności. Nie wiem czy dobrze je zrobiłam, więc chciałabym prosić o sprawdzenie
1)
\(\displaystyle{ (2cosx - 1)\cdot (\frac{\sqrt{2}}{2} -sinx)=0}\)
\(\displaystyle{ 2cosx-1=0 \frac{\sqrt{2}}{2} -sinx=0}\)
\(\displaystyle{ cosx= \frac{1}{2} sinx= \frac{\sqrt{2}}{2}}\)
\(\displaystyle{ cosx=cos \frac{\pi}{3} sinx=sin \frac{\pi}{4}}\)
\(\displaystyle{ x=\frac{\pi}{3}+k2\pi x=-\frac{\pi}{3}+k2\pi x=\frac{\pi}{4}+k2\pi x=\pi-\frac{\pi}{4}+k2\pi}\)
\(\displaystyle{ x=\frac{\pi}{3}+k2\pi x=-\frac{\pi}{3}+k2\pi x=\frac{\pi}{4}+k2\pi x=\frac{3\pi}{4}+k2\pi}\)
\(\displaystyle{ k\in C}\)
2)
\(\displaystyle{ tg^{2} x qslant 3}\)
\(\displaystyle{ tgx qslant \sqrt{3} tgx qslant -\sqrt{3}}\)
\(\displaystyle{ tgx=tg\frac{\pi}{3} tgx=tg(-\frac{\pi}{3}}\)
\(\displaystyle{ x=\frac{\pi}{3}+k\pi x=-\frac{\pi}{3}+k\pi}\)
\(\displaystyle{ x\in , k\in C}\)
3)
\(\displaystyle{ ctg^{2}x< \frac{1}{3}}\)
\(\displaystyle{ ctgx< \frac{\sqrt{3}}{3} \vee ctgx > -\frac{\sqrt{3}}{3}}\)
\(\displaystyle{ ctgx=ctg \frac{\pi}{3} \vee ctgx=ctg(-\frac{\pi}{3}}\)
\(\displaystyle{ x=\frac{\pi}{3}+k\pi \vee x=-\frac{\pi}{3}+k\pi}\)
\(\displaystyle{ x\in (-\frac{\pi}{3}+k\pi ; \frac{\pi}{3}+k\pi) , k\in C}\)
4)
\(\displaystyle{ sin^{2}x- \frac{1}{2}>0}\)
\(\displaystyle{ sinx> \frac{\sqrt{2}}{2} sinx< -\frac{\sqrt{2}{2}}\)
\(\displaystyle{ sinx=sin \frac{\pi}{4} sinx=sin(-\frac{\pi}{4}}\)
\(\displaystyle{ x=\frac{\pi}{4}+k2\pi x=\pi -\frac{\pi}{4}+k2\pi x=-\frac{\pi}{4}+k2\pi x=\pi+\frac{\pi}{4}+k2\pi , k\in C}\)
\(\displaystyle{ x=\frac{\pi}{4}+k2\pi x=\frac{3\pi}{4}+k2\pi x=-\frac{\pi}{4}+k2\pi x=\frac{5\pi}{4}+k2\pi , k\in C}\)
