Skoro

$x\ne \frac{\pi }{2}+k\pi$  

to

$\cos x\ne 0$  

 

a) Ze wzoru na sinus różnicy:

$\sin x\cdot \cos \left(\frac{\pi }{6}\right)-\sin \left(\frac{\pi }{6}\right)\cdot \cos x=\frac{2}{3}\cos x$  

$\sin x\cdot \frac{\sqrt{3}}{2}-\frac{1}{2}\cos x=\frac{2}{3}\cos x$  

$\frac{\sqrt{3}}{2}\sin x=\frac{1}{2}\cos x+\frac{2}{3}\cos x$  

$\frac{\sqrt{3}}{2}\sin x=\frac{3}{6}\cos x+\frac{4}{6}\cos x$  

$\frac{\sqrt{3}}{2}\sin x=\frac{7}{6}\cos x\mid :\cos x$   

$\frac{\sqrt{3}}{2}\frac{\sin x}{\cos x}=\frac{7}{6}\mid \cdot \frac{2}{\sqrt{3}}$  

$\mathrm{tg}x=\frac{7}{6}\cdot \frac{2}{\sqrt{3}}$  

$\mathrm{tg}x=\frac{14}{6\sqrt{3}}$  

$\mathrm{tg}x=\frac{14\sqrt{3}}{6\cdot 3}$  

$\mathrm{tg}x=\frac{7\sqrt{3}}{9}$  

 

$b)\cos x\cdot \cos \left(x+\frac{\pi }{4}\right)=\sin 2x$  

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

$\cos x\left(\cos x\cdot \frac{\sqrt{2}}{2}-\sin x\cdot \frac{\sqrt{2}}{2}\right)=\sin 2x$  

$\frac{\sqrt{2}}{2}\cos x\left(\cos x-\sin x\right)=2\sin x\cos x$  

$\frac{\sqrt{2}}{2}{\cos }^{2}x-\frac{\sqrt{2}}{2}\sin x\cos x-2\sin x\cos x=0\mid \cdot 2$  

$\sqrt{2}{\cos }^{2}x-\sqrt{2}\sin x\cos x-4\sin x\cos x=0$  

$\sqrt{2}{\cos }^{2}x-\left(\sqrt{2}+4\right)\sin x\cos x=0$  

$\cos x\cdot \left[\sqrt{2}\cos x-\left(\sqrt{2}+4\right)\sin x\right]=0\mid :\cos x$  

$\sqrt{2}\cos x-\left(\sqrt{2}+4\right)\sin x=0$  

$\sqrt{2}\cos x=\left(\sqrt{2}+4\right)\sin x$  

$\frac{\sin x}{\cos x}=\frac{\sqrt{2}}{\sqrt{2}+4}$  

Usuńmy niewymierność z mianownika:

$\frac{\sqrt{2}}{\sqrt{2}+4}\cdot \frac{\sqrt{2}-4}{\sqrt{2}-4}=\frac{\sqrt{2}\left(\sqrt{2}-4\right)}{2-16}=\frac{\sqrt{2}\left(\sqrt{2}-4\right)}{-14}=\frac{\sqrt{2}\left(4-\sqrt{2}\right)}{14}=\frac{2\left(2\sqrt{2}-1\right)}{14}=\frac{2\sqrt{2}-1}{7}$   

czyli

$\mathrm{tg}x=\frac{2\sqrt{2}-1}{7}$