Question

1)sin x/2=1 2)cos 2x +2=0 3)sin 2x=-1/2 4)2cos x +√3=0 5)√3 tg 2x+1=0 Решите задачи паже))

Answer 1

Решение во вложении..............

Answer 2

1) sin(x/2) = 1

To solve the equation sin(x/2) = 1, we can start by finding the values of x that satisfy this equation.

Using the trigonometric identity sin(x/2) = ±√[(1 - cos(x))/2], we can rewrite the equation as:

√[(1 - cos(x))/2] = 1

Squaring both sides of the equation, we get:

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

Multiplying both sides by 2, we have:

1 - cos(x) = 2

Rearranging the equation, we get:

cos(x) = -1

The cosine function equals -1 at π radians (180 degrees) and its multiples. Therefore, the solutions to the equation sin(x/2) = 1 are:

x/2 = π + 2πn, where n is an integer

Solving for x, we have:

x = 2π + 4πn, where n is an integer

So, the solutions to the equation sin(x/2) = 1 are x = 2π + 4πn, where n is an integer.

2) cos(2x) + 2 = 0

To solve the equation cos(2x) + 2 = 0, we can start by isolating the cosine term.

Subtracting 2 from both sides of the equation, we get:

cos(2x) = -2

Since the cosine function has a range of -1 to 1, there are no real solutions to the equation cos(2x) + 2 = 0.

3) sin(2x) = -1/2

To solve the equation sin(2x) = -1/2, we can use the inverse sine function to find the values of x.

Taking the inverse sine of both sides of the equation, we have:

2x = arcsin(-1/2)

Using the unit circle or a calculator, we find that the principal value of arcsin(-1/2) is -π/6.

Therefore, the solutions to the equation sin(2x) = -1/2 are:

2x = -π/6 + 2πn, where n is an integer

Solving for x, we have:

x = -π/12 + πn, where n is an integer

So, the solutions to the equation sin(2x) = -1/2 are x = -π/12 + πn, where n is an integer.

4) 2cos(x) + √3 = 0

To solve the equation 2cos(x) + √3 = 0, we can start by isolating the cosine term.

Subtracting √3 from both sides of the equation, we get:

2cos(x) = -√3

Dividing both sides by 2, we have:

cos(x) = -√3/2

Using the unit circle or a calculator, we find that the principal value of cos(x) = -√3/2 is 5π/6.

Therefore, the solutions to the equation 2cos(x) + √3 = 0 are:

x = π ± 2πn, where n is an integer

So, the solutions to the equation 2cos(x) + √3 = 0 are x = π ± 2πn, where n is an integer.

5) √3 + tg(2x) + 1 = 0

To solve the equation √3 + tg(2x) + 1 = 0, we can start by isolating the tangent term.

Subtracting √3 and 1 from both sides of the equation, we get:

tg(2x) = -√3 - 1

Taking the inverse tangent of both sides of the equation, we have:

2x = arctan(-√3 - 1)

Using a calculator, we find that the principal value of arctan(-√3 - 1) is approximately -1.107.

Therefore, the solutions to the equation √3 + tg(2x) + 1 = 0 are:

2x = -1.107 + πn, where n is an integer

Solving for x, we have:

x = -0.554 + πn, where n is an integer

So, the solutions to the equation √3 + tg(2x) + 1 = 0 are x = -0.554 + πn, where n is an integer.

I hope this helps! Let me know if you have any further questions.