Question

Ctg2x+4=0 Cos3x=-1/2 2sinx+√3=0 Cosx/2=0 4sin2xcos2x=√3

Answer 1

Ответ:

Пошаговое объяснение:

ctg2x=-4,  2x=p-arctg4+pn,  x=p/2-1/2 arctg4+pn/2

3x=+-2p/3+2pn,  x=+- 2p/9+2pn/3

snx=-V3/2,  x=(-1)^n*(-p/3)+pn

x/2=p/2+pn, x=p+2pn

sin4x=V3/2,    4x=(-1)^n*p/3+pn,  x=(-1)^n*p/12+pn/4,  n  E  Z

Answer 2

Let's solve each of the equations step by step:

1. Ctg(2x) + 4 = 0:

To solve this equation, we'll rearrange it to isolate the cotangent term: Ctg(2x) = -4

Now, we can find the angle whose cotangent is -4. Since cotangent is the reciprocal of tangent, we can write: Tan(2x) = -1/4

Using the inverse tangent function, we find: 2x = arctan(-1/4)

However, the tangent function has a periodicity of π, so we need to consider multiple solutions: 2x = arctan(-1/4) + πn, where n is an integer

Finally, we can solve for x: x = (arctan(-1/4) + πn)/2, where n is an integer

2. Cos(3x) = -1/2:

To solve this equation, we need to find the angles whose cosine is -1/2. This can be done using inverse cosine: 3x = arccos(-1/2)

Since cosine is periodic with a period of 2π, we need to consider multiple solutions: 3x = arccos(-1/2) + 2πn, where n is an integer

Finally, we can solve for x: x = (arccos(-1/2) + 2πn)/3, where n is an integer

3. 2sin(x) + √3 = 0:

To solve this equation, we'll isolate the sine term: 2sin(x) = -√3

Dividing both sides by 2, we get: sin(x) = -√3/2

Now, we find the angles whose sine is -√3/2 using inverse sine: x = arcsin(-√3/2)

The sine function has a periodicity of 2π, so we need to consider multiple solutions: x = arcsin(-√3/2) + 2πn, where n is an integer

4. Cos(x/2) = 0:

To solve this equation, we can use inverse cosine: x/2 = arccos(0)

Since the cosine of 0 is 1 at multiples of π, we have: x/2 = πn, where n is an integer

Multiplying both sides by 2, we find: x = 2πn, where n is an integer

5. 4sin(2x)cos(2x) = √3:

To solve this equation, we'll use the double-angle formula for sine: 2sin(2x)cos(2x) = √3

Dividing both sides by 2, we get: sin(2x)cos(2x) = √3/2

Now, we can rewrite the left side using the sine double-angle formula: sin(4x)/2 = √3/2

Multiplying both sides by 2, we obtain: sin(4x) = √3

To find the angles whose sine is √3, we can use inverse sine: 4x = arcsin(√3)

The sine function has a periodicity of 2π, so we need to consider multiple solutions: 4x = arcsin(√3) + 2πn, where n is an integer

Finally, we can solve for x: x = (arcsin(√3) + 2πn)/4, where n is an integer

Please note that the solutions provided are general solutions that consider the periodicity of trigonometric functions.