Question

1)Cos X = -(√3/2) 2)sin x =√2/2 3)cos3x = √3/2 4)cos(x/2)= √3/2 5) sin(x/2)=√2/2 6)sin3x=√2/2 7)sin(x/3)=-√2

Answer 1

1
x=+-5π/6+2πn,n∈z
2
x=(-1)^n*π/4+πn,n∈z
3
3x=+-π/6+2πn
x=+-π/18+2πn/3
4
x/2=+-π/6+2πn
x=+-π/3+πn,n∈z
5
x/2=(-1)^n*π/4+πn
x=(-1)^n*π/2+2πn,n∈z
6
3x=(-1)^n*π/4+πn
x=(-1)^n*π/12+πn/3,n∈z
7
sin(x/3)=-√2<-1
нет решения

Answer 2

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Answer 3

To solve these trigonometric equations, we can use the values of trigonometric functions at specific angles. Here are the solutions for each equation:

1. Cos X = -(√3/2): The angle X is 150 degrees or 5π/6 radians. This is based on the unit circle, where the cosine function is negative in the second and third quadrants.

2. Sin x = √2/2: The angle x is 45 degrees or π/4 radians. This is a well-known angle where both sine and cosine functions are equal to 1/√2.

3. Cos 3x = √3/2: To find x, we need to solve for 3x first. Taking the inverse cosine of √3/2, we get 3x = 30 degrees or π/6 radians. Dividing by 3, we find x = 10 degrees or π/18 radians.

4. Cos(x/2) = √3/2: To find x, we can double the angle from equation 3). Therefore, x = 20 degrees or π/9 radians.

5. Sin(x/2) = √2/2: Similar to equation 2), x/2 is 45 degrees or π/4 radians. Therefore, x = 90 degrees or π/2 radians.

6. Sin 3x = √2/2: Again, we need to solve for 3x first. Taking the inverse sine of √2/2, we get 3x = 45 degrees or π/4 radians. Dividing by 3, we find x = 15 degrees or π/12 radians.

7. Sin(x/3) = -√2: To find x, we can multiply the angle from equation 6) by 3. Therefore, x = 45 degrees or π/4 radians.

Please note that these are the principal solutions, and trigonometric functions have periodicity, so there are infinitely many solutions to each equation.