Question

Помогите решить второй вариант!
Хотя бы 2-3 задания

Answer 1

№1.

$cos(x)=0\\x=\frac{\pi}{2} +\pi *n$ , n∈Z.

Ответ: б.

№2.

а)

$tg3x=-\frac{\sqrt{3}}{3} \\tg3x=-\frac{1}{\sqrt{3}} \\3x=-\frac{\pi}{6} +\pi *n\\x=-\frac{\pi}{18} +\frac{\pi *n}{3}$

, n∈Z.

б)

$sin(\frac{x}{2}-\frac{\pi}{3})+\frac{\sqrt{3}}{2}=0\\-cos(\frac{\pi}{2}+\frac{x}{2}-\frac{\pi}{3})=-\frac{\sqrt{3}}{2}\\cos(\frac{x}{2}+\frac{\pi}{6})=\frac{\sqrt{3}}{2}$

$\frac{x}{2}+\frac{\pi}{6} = 2\pi*n_{1} +\frac{\pi}{6}$ , n1∈Z;

$\frac{x}{2}+\frac{\pi}{6} = 2\pi*n_{2} +\frac{11\pi}{6}$ , n2∈Z.

$\frac{x}{2} = 2\pi*n_{1}$ , n1∈Z;

$\frac{x}{2}= 2\pi*n_{2} +\frac{5\pi}{3}$ , n2∈Z.

$x = 4\pi*n_{1}$ , n1∈Z;

$x = 4\pi*n_{2} +\frac{10\pi }{3}$ , n2∈Z.

№3.

$\sqrt{3} *cos(x) + sin(2x)=0;\\\sqrt{3} *cos(x) + 2*sin(x)cos(x)=0, | :cos(x);\\\sqrt{3} + 2*sin(x)=0;\\2*sin(x)=\sqrt{3};\\sin(x)=\frac{\sqrt{3}}{2} .$

$x =\frac{4\pi }{3}+2\pi*n_{1}$ , n1∈Z;

$x =\frac{5\pi }{3}+2\pi*n_{2}$ , n2∈Z.

№4.

а)

$cos^{2}(2x)-2cos(2x)+1=0;\\(cos(2x)-1)^2=0;\\cos(2x)-1=0;\\cos(2x)=1.$

$2x=2\pi *n$ , n∈Z;

$x=\pi *n$ , n∈Z.

б)

$cos(9x)-cos(5x)-\sqrt{3}*sin(2x)=0;\\-2*sin(\frac{9x-5x}{2})*sin(\frac{9x+5x}{2} )-\sqrt{3}*sin(2x)=0;\\-2*sin(2x)*sin(7x)-\sqrt{3}*sin(2x)=0;\\-sin(2x)*(2*sin(7x)+\sqrt{3})=0;\\sin(2x)*(2*sin(7x)+\sqrt{3})=0.$

1.

$sin(2x)=0;$

$2x=\pi *n$ , n∈Z;

$x=\frac{\pi *n}{2}$ , n∈Z.

2.

$2*sin(7x)+\sqrt{3}=0;\\2*sin(7x)=-\sqrt{3};\\sin(7x)=-\frac{\sqrt{3}}{2} .$

$7x=\frac{4\pi }{3} +2\pi *n_{1}$ , n1∈Z;

$x=\frac{4\pi }{21} +\frac{2\pi *n_{1}}{7}$ , n1∈Z;

$7x=\frac{5\pi }{3} +2\pi *n_{2}$ , n2∈Z;

$x=\frac{5\pi }{21} +\frac{2\pi *n_{2}}{7}$ , n2∈Z.

в)

$2cos^2(x)-3*sin(x)*cos(x)+sin^2(x)=0,\\cos^2(x)-sin(x)*cos(x)+cos^2(x)-2*sin(x)*cos(x)+sin^2(x)=0,\\cos(x)*(cos(x)-sin(x))+(cos(x)-sin(x))^2=0;\\cos(x)*(cos(x)-sin(x))+(cos(x)-sin(x))*(cos(x)-sin(x))=0;\\(cos(x)-sin(x))*(2cos(x)-sin(x))=0.$

1.

$cos(x)-sin(x)=0;\\cos(x)=sin(x), |:sin(x);\\ctg(x)=1.$

$x=\frac{\pi }{4} +\pi *n_{1}$ , n1∈Z.

2.

$2cos(x)-sin(x)=0;\\2cos(x)=sin(x), |:sin(x);\\2tg(x)=1\\ctg(x)=\frac{1}{2} .$

$x=arctg\frac{1 }{2} +\pi *n_{2}$ , n2∈Z.

№5.

$arccos(x) = \pi,\\x=-1.$

$arccos(13-9x+x^2)=\pi ,\\13-9x+x^2=-1,\\x^2-9x+14=0,\\x^2-7x-2x+14=0,\\x*(x-7)-2*(x-7)=0,\\(x-7)*(x-2)=0.$

$x-7=0;\\x=7.$

$x-2=0;\\x=2.$