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Question 1

Question 2

$1. a) \frac{1}{2}arccos(-\frac{\sqrt3}{2})-4arcsin(-\frac{\sqrt2}{2})=\frac{1}{2}(\pi-arccos(\frac{\sqrt3}{2}))+\\+4arcsin(\frac{\sqrt2}{2})=\frac{1}{2}(\pi-\frac{\pi}{6})+4\frac{\pi}{4}=\frac{5\pi}{12}+\pi=\frac{17\pi}{12}\\b) tg(5arctg\frac{\sqrt3}{3}-\frac{1}{4}arcsin\frac{\sqrt3}{2})=tg(5\frac{\pi}{6}-\frac{1}{4}\frac{\pi}{3})=tg(\frac{5\pi}{6}-\frac{\pi}{12})=\\=tg(\frac{9\pi}{12})=tg(\frac{\pi}{4})=1$
$2. a) 3sin^2 2x+2sin2x-1=0\\sin2x=t, -1\leq t\leq 1\\3t^2+2t-1=0\\D=4+12=16\\t_1=\frac{-2-4}{6}=-1 \\sin2x=-1\\ 2x=\frac{3\pi}{2}+2\pi n, n\in Z\\x_1=\frac{3\pi}{4}+\pi n, n\in Z \\\\t_2=\frac{-2+4}{6}=\frac{1}{3}\\sin2x=\frac{1}{3}\\2x=arcsin\frac{1}{3}+2\pi k, k\in Z\\x_2=\frac{1}{2}arcsin\frac{1}{3}+\pi k, k\in Z$
$b) 4sin^2x+sinx *cos x-3cos^2 x=0\\4\frac{1-cos2x}{2}+\frac{1}{2}sin2x-3\frac{1+cos2x}{2}=0\\4-4cos2x+sin2x-3-3cos2x=0\\1-7cos2x+sin2x=0\\\sqrt{1-cos^22x}=7cos2x-1\\1-cos^22x=49cos^22x-14cos2x+1\\50cos^22x-14cos2x=0\\cos2x(25cos2x-7)=0\\cos2x_1=0 , cos2x_2=\frac{7}{25}\\2x_1=\frac{\pi}{2}+\pi n, n\in Z, x_1=\frac{\pi}{4}+\frac{\pi n}{2}, n\in Z\\2x_2=arccos\frac{7}{25}+2\pi k, k\in Z, x_2=\frac{1}{2}arccos\frac{7}{25}+\pi k, k\in Z$

$3. [tex]sin(\frac{4x}{3}+\frac{\pi}{6})=-\frac{1}{2}\\I)\frac{4x}{3}+\frac{\pi}{6}=\frac{7\pi}{6}+2\pi n, n\in Z\\\frac{4x}{3}=\pi +2\pi n, n\in Z\\x=\frac{3\pi}{4}+\frac{3\pi}{2}n,n\in Z\\x\in[-2\pi; 2\pi]\\-2\pi \leq\frac{3\pi}{4}+\frac{3\pi}{2}n \leq2\pi\\-8\leq 3+6n\leq 8\\-11\leq 6n \leq 5\\-\frac{11}{6}\leq n \leq \frac{5}{6}\\n=-1; 0\\n=-1=\ \textgreater \ x=\frac{3\pi}{4}-\frac{3\pi}{2}=-\frac{3\pi}{2}\\n=0=\ \textgreater \ \frac{3\pi}{4}$

$\\II)\frac{4x}{3}+\frac{\pi}{6}=-\frac{\pi}{6}+2\pi k, k\in Z\\\frac{4x}{3}=-\frac{\pi}{3}+2\pi k, k\in Z\\x=\frac{\pi}{4}+\frac{3\pi}{2}k, k\in Z\\x\in [-2\pi; 2\pi]\\-2\pi \leq \frac{\pi}{4}+\frac{3\pi}{2}k\leq 2\pi\\-8\leq 1+6k\leq8\\-9\leq 6k \leq 7\\-\frac{3}{2}\leq k\leq \frac{7}{6}\\k=-1; 0; 1\\k=-1=\ \textgreater \ x=\frac{\pi}{4}-\frac{3\pi}{2}=-\frac{\pi}{4}\\k=0=\ \textgreater \ x=\frac{\pi}{4}\\k=1=\ \textgreater \ x=\frac{7\pi}{4}$

$4. \left \{ {{sinx\leq \frac{\sqrt3}{2}} \atop {cos x \ \textgreater \ -\frac{1}{7}}} \right. \\arccos\frac{1}{7}-\pi\ \textless \ x\leq \frac{\pi}{3}$
Решение - графически, см. в приложении.

$5. arcsin(\sqrt{x-5})=arcsin(3-\sqrt{10-x})\\D:\\ \left \{ {{\sqrt{x-5}\leq 1} \atop {-1 \leq 3-\sqrt{10-x}\leq 1}} \right. , \left \{ {{5\leq x\leq 6} \atop {-4\leq -\sqrt{10-x}\leq -2}} \right. , \left \{ {{5\leq x\leq 6} \atop {2\leq \sqrt{10-x}\leq 4}} \right. , \left \{ {{5\leq x\leq 6} \atop {4\leq {10-x}\leq 16}} \right. \\\left \{ {{5\leq x\leq 6} \atop {-6\leq x\leq 6}} \right. 5\leq x\leq 6\\ \\\sqrt{x-5}=3-\sqrt{10-x}$
$\sqrt{x-5}=3-\sqrt{10-x}\\\sqrt{x-5}+\sqrt{10-x}=3\\x-5+10-x+2\sqrt{(x-5)(10-x)}=9\\\sqrt{(x-5)(10-x)}=2\\(x-5)(10-x)=4\\10x-x^2-50+5x-4=0\\x^2-15x+54=0\\x_1=7, x_2=8\\D: 5\leq x\leq 6\\NoSolutions$