$a\text{)}\sin \left(\frac{\pi }{4}-x\right)=\sin 2x$  

Zatem:

$\frac{\pi }{4}-x=2x+2k\pi \text{lub}\frac{\pi }{4}-x=\pi -2x+2k\pi$  

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

$x=\frac{\pi }{12}+\frac{2k\pi }{3}\text{lub}x=\frac{3\pi }{4}+2k\pi ,k\in \mathbb{Z}$  

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$b\text{)}\cos \left(3x-\frac{\pi }{3}\right)=\cos x$  

Zatem:

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

$2x-\frac{\pi }{3}=2k\pi \text{lub}4x-\frac{\pi }{3}=2k\pi$  

$2x=\frac{\pi }{3}+2k\pi \text{lub}4x=\frac{\pi }{3}+2k\pi$  

$x=\frac{\pi }{6}+k\pi \text{lub}x=\frac{\pi }{12}+\frac{k\pi }{2},k\in \mathbb{Z}$  

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$c\text{)}\mathrm{tg}\left(\frac{\pi }{6}-x\right)=\mathrm{tg}\left(2x+\frac{\pi }{4}\right)$  

Określmy dziedzinę tego równania:

$\frac{\pi }{6}-x\ne \frac{\pi }{2}+k\pi \text{i}x\ne 2x+\frac{\pi }{4}+k\pi$  

$-x\ne \frac{\pi }{3}+k\pi \text{i}-x\ne \frac{\pi }{4}+k\pi$  

Stąd:

$x\ne -\frac{\pi }{3}+k\pi \text{i}x\ne -\frac{\pi }{4}+k\pi ,k\in \mathbb{Z}$  

 

Zatem:

$\frac{\pi }{6}-x=2x+\frac{\pi }{4}+k\pi ,k\in \mathbb{Z}$  

$\frac{\pi }{6}-3x=\frac{\pi }{4}+k\pi ,k\in \mathbb{Z}$  

$-3x=\frac{\pi }{12}+k\pi ,k\in \mathbb{Z}$  

$x=-\frac{\pi }{36}+\frac{k\pi }{3},k\in \mathbb{Z}$  

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$d\text{)}\sin \left(x-\frac{2}{3}\pi \right)=\cos x$  

$\sin \left(x-\frac{2}{3}\pi \right)=\sin \left(\frac{\pi }{2}-x\right)$  

Zatem:

$x-\frac{2}{3}\pi =\frac{\pi }{2}-x+2k\pi \text{lub}x-\frac{2}{3}\pi =\pi -\left(\frac{\pi }{2}-x\right)+2k\pi$  

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

$2x=\frac{7\pi }{6}+2k\pi \text{lub}-\frac{2}{3}\pi =\pi -\frac{\pi }{2}+2k\pi$  

Zauważmy, że drugie równanie jest sprzeczne, więc:

$x=\frac{7\pi }{12}+k\pi ,k\in \mathbb{Z}$  

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$e\text{)}\cos 2x=\sin \left(\frac{\pi }{12}-x\right)$  

Zatem:

$\sin \left(\frac{\pi }{2}-2x\right)=\sin \left(\frac{\pi }{12}-x\right)$  

$\frac{\pi }{2}-2x=\frac{\pi }{12}-x+2k\pi \text{lub}\frac{\pi }{2}-2x=\pi -\frac{\pi }{12}-x+2k\pi$  

$-x+\frac{\pi }{2}=\frac{\pi }{12}+2k\pi \text{lub}-3x+\frac{\pi }{2}=\frac{11\pi }{12}+2k\pi$  

$-x=-\frac{5\pi }{12}+2k\pi \text{lub}-3x=\frac{5\pi }{12}+2k\pi$  

$x=\frac{5\pi }{12}+2k\pi \text{lub}x=-\frac{5\pi }{36}+\frac{2k\pi }{3}$  

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$f\text{)}\mathrm{tg}\left(\frac{\pi }{2}+\frac{\pi }{3}\right)=\mathrm{tg}\left(\frac{x}{3}-\frac{\pi }{4}\right)$

Określmy dziedzinę tego równania:

$x\ne \frac{x}{3}-\frac{\pi }{4}+k\pi$  

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

Stąd:

$x\ne -\frac{3\pi }{8}+\frac{3k\pi }{2},k\in \mathbb{Z}$  

 

Zatem: 

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

$\frac{5\pi }{6}=\frac{x}{3}-\frac{\pi }{4}+k\pi$  

$-\frac{x}{3}+\frac{10\pi }{12}=-\frac{3\pi }{12}+k\pi$  

$-\frac{x}{3}=-\frac{13\pi }{12}+k\pi$  

$x=\frac{13\pi }{4}+3k\pi ,k\in \mathbb{Z}$