Question

Решите пожалуйста срочно очень прошу

Answer 1

$1.\\\frac{7\pi}6=\frac{7\cdot180}{6}=7\cdot30=210^o\\\frac{11\pi}5=\frac{11\cdot180}5=11\cdot36=396^o\\\\2.\\330^o=\frac{330\pi}{180}=\frac{11\pi}6\\\\120^o=\frac{120\pi}{180}=\frac{2\pi}3\\\\3.\;160^o\;-\;II\;4ETB.,\sin160^o>0\\tg\left(-\frac{13\pi}6\right)=-tg(390^o)<0\\\\4.\;\cos765^o=\cos\frac{17\pi}4=\cos(2\pi+\frac\pi4)=\cos\frac\pi4=\frac1{\sqrt2}\\ctg\frac{19\pi}4=ctg\left(4\pi+\frac{3\pi}4\right)=ctg\frac{3\pi}4=-1$

$5.\;3tg\frac\pi5\cdot ctg\frac\pi5+2\sin\frac\pi3=3\cdot1+3\cdot\frac{\sqrt3}2=3+\frac{3\sqrt3}2\\\\6.\;\frac{2\sin\left(\frac\pi2-\alpha\right)}{5\cos(\pi+\alpha)}=\frac{2\cos\alpha}{-5\cos\alpha}-\frac25\\\\7.\;2\cos x+\sqrt3=0\\\cos x=-\frac{\sqrt3}2\\x=\pm\frac{5\pi}6+\pi n,\;n\in\mathbb{Z}\\\\2\sin\left(\frac\pi3-x\right)=1\\\sin\left(\frac\pi3-x\right)=\frac12\\\frac\pi3-x=(-1)^n\cdot\frac\pi6+\pi n\\x=\frac\pi3-(-1)^n\cdot\frac\pi6-\pi n,\;n\in\mathbb{Z}$