Question

Решите уравнения:1) 2cos10x=12) sin0,2x =-13) sin2x - sinx=0

Answer 1

1) cos10x=1/2

10x=a       cos a=1/2   a=±arccos(1/2)+2*\pi*n=±\pi/3+2*\pi*n n∈Z

a=10x=±\pi/3+2*\pi*n                     x=±\pi/30+\pi*n/5 n∈Z

2) sin 0.2x=-1    a=0.2x

sin a =-1      a= - pi/2+2*pi*n ,     a= 3/2*pi+2*pi*n     n∈Z

0.2x= -pi/2+2*pi*n   x= -5/2*pi+10*pi*n    n∈Z

0.2x=3/2*pi+2*pi*n   x=15/2*pi+10*pi*n    n∈Z

3)  sin 2x - sin x=0    

2* sin x * cos x - sin x =0       sin x  (2* cos x - 1)=0      

3.1)     sin x =0   x=pi*n    n∈Z  

3.2)    2cos x =1      cos x =1/2         x=±pi/3+2*pi*n     n∈Z

Answer 2

1. 2cos(10x) = 1

To solve this equation, we can start by isolating cos(10x) on one side:

2cos(10x) = 1 cos(10x) = 1/2

Now, we need to find the values of x that satisfy this equation. Since the cosine function has a period of 2π, we can use the inverse cosine function (arccos) to find the solutions within one period. The general solution is given by:

10x = arccos(1/2) + 2πn, where n is an integer

We can simplify the equation further:

10x = π/3 + 2πn

Dividing both sides by 10:

x = (π/30) + (2πn)/10 x = π/30 + πn/5

So, the general solution for the equation 2cos(10x) = 1 is:

x = π/30 + πn/5, where n is an integer.

2. sin(0.2x) = -1

To solve this equation, we'll isolate sin(0.2x) on one side:

sin(0.2x) = -1

Since the sine function has a period of 2π, we can use the inverse sine function (arcsin) to find the solutions within one period. The general solution is given by:

0.2x = arcsin(-1) + 2πn, where n is an integer

Simplifying the equation:

0.2x = -π/2 + 2πn

Dividing both sides by 0.2:

x = (-π/2)/0.2 + (2πn)/0.2 x = -2.5π + 10πn

So, the general solution for the equation sin(0.2x) = -1 is:

x = -2.5π + 10πn, where n is an integer.

3. sin(2x) - sin(x) = 0

To solve this equation, we'll first combine the terms on one side:

sin(2x) - sin(x) = 0

Using the identity sin(A) - sin(B) = 2sin((A-B)/2)cos((A+B)/2), we can rewrite the equation as:

2sin((2x - x)/2)cos((2x + x)/2) = 0

Simplifying further:

2sin(x/2)cos(3x/2) = 0

Now, we have two cases to consider:

Case 1: sin(x/2) = 0

For this case, we have:

x/2 = 0 + πn, where n is an integer

Simplifying:

x = 2πn, where n is an integer

Case 2: cos(3x/2) = 0

For this case, we have:

3x/2 = π/2 + πn, where n is an integer

Simplifying:

x = (2π/3) + (2πn/3), where n is an integer

So, the general solution for the equation sin(2x) - sin(x) = 0 is:

x = 2πn or x = (2π/3) + (2πn/3), where n is an integer.