РЕШИТЕ ПОЖАЛУЙСТА ХОТЯ БЫ НЕСКОЛЬКО,ПОЖАЛУЙСТА

Решите задачу:

$1) \ x^2=6^2+(3\sqrt2)^2-2\cdot6\cdot3\sqrt2\cos45^\circ=36+9\cdot2-36\sqrt2\cdot\frac{\sqrt2}{2}=\\=36+18-36=18, \\ x=\sqrt{18}=3\sqrt2;$

$2) \ x^2=4^2+3^2-2\cdot4\cdot3\cos(180^\circ-60^\circ)=16+9+24\cos60^\circ=\\=25+24\cdot\frac{1}{2}=25+12=37, \\ x=\sqrt{37};$

$3) \ \frac{x}{\sin(180^\circ-(15^\circ+135^\circ))}=\frac{\sqrt2}{\sin135^\circ}, \\ x=\frac{\sqrt2\sin(180^\circ-150^\circ)}{\sin(180^\circ-45^\circ)}=\frac{\sqrt2\sin30^\circ}{\sin45^\circ}=\frac{\sqrt2\cdot\frac{1}{2}}{\frac{\sqrt2}{2}}=1;$

$4) \ \frac{x}{\sin45^\circ}=\frac{\sqrt3}{\sin(180^\circ-(15^\circ+45^\circ))}, \\ x=\frac{\sqrt3\sin45^\circ}{\sin(180^\circ-60^\circ)}=\frac{\sqrt3\sin45^\circ}{\sin60^\circ}=\frac{\sqrt3\cdot\frac{\sqrt2}{2}}{\frac{\sqrt3}{2}}=\sqrt2;$

$5) \ \frac{x}{\sin(180^\circ-(\alpha+\beta))}=\frac{a}{\sin\alpha}, \\ x=\frac{a\sin(\alpha+\beta)}{\sin\alpha};$

$6) \ a^2=b^2+c^2-2bc\cos x, \\ \cos x= \frac{b^2+c^2-a^2}{2bc}.$