Question

1)cos2x+2=0. 2)sin4x=0 3)2sinx/2+1=0 4)2cos2x-1=0 5)tg^2x-tgx=0

Answer 1

1) cos2x=-2

нет решений

-1<=cos2x<=1

 

2) sin4x=0

arcsin0=0

4x = пи*n

x= $\frac{\pi\cdot n}{4}$

n пренадлежит Z

 

3) sinx/2=-1/2

 arcsin -1/2 =  - пи/6

$\frac{x}{2}=(-1)^{n+1}\cdot \frac{\pi}{6} + \pi\cdot n$

$x=(-1)^{n+1}\cdot \frac{\pi}{3} +2\cdot \pi\cdot n$

n пренадлежит Z

 

4) )cos2x=1/2

arccos 1/2 = пи/3

2x= +- пи/3 + 2*пи*n

x= +- пи/6 + пи*n

n пренадлежит Z

 

5) tg^2x=tgx

tgx=0

arctg0=0

x=пи*n 

n пренадлежит Z

Answer 2

Sure! Let's go through each of the equations one by one and find the solutions.

1) cos(2x) + 2 = 0:

To solve this equation, we'll first isolate the cosine term. Subtracting 2 from both sides, we get:

cos(2x) = -2

Now, we need to find the values of x that satisfy this equation. However, the cosine function ranges between -1 and 1, so there are no real solutions for this equation. In other words, there are no values of x that make cos(2x) equal to -2.

2) sin(4x) = 0:

To solve this equation, we'll use the fact that the sine function is equal to zero at multiples of π. So, we can set 4x equal to nπ, where n is an integer:

4x = nπ

Now, we'll solve for x by dividing both sides by 4:

x = nπ/4

This equation gives us infinitely many solutions, where n is an integer. Each solution corresponds to a value of x for which sin(4x) is equal to zero.

3) 2sin(x/2) + 1 = 0:

To solve this equation, we'll first isolate the sine term. Subtracting 1 from both sides, we get:

2sin(x/2) = -1

Next, we'll divide both sides by 2:

sin(x/2) = -1/2

Now, we need to find the values of x/2 that satisfy this equation. To do that, we can check the unit circle or use the inverse sine function. Since -1/2 corresponds to an angle of -π/6 or 7π/6 on the unit circle, we can write:

x/2 = -π/6 + 2πn or x/2 = 7π/6 + 2πn

Solving for x, we multiply both sides by 2:

x = -π/3 + 4πn or x = 7π/3 + 4πn

Here, n is an integer, and each solution corresponds to a value of x for which 2sin(x/2) + 1 is equal to zero.

4) 2cos(2x) - 1 = 0:

To solve this equation, we'll first isolate the cosine term. Adding 1 to both sides, we get:

2cos(2x) = 1

Next, we'll divide both sides by 2:

cos(2x) = 1/2

Now, we need to find the values of 2x that satisfy this equation. To do that, we can check the unit circle or use the inverse cosine function. Since 1/2 corresponds to an angle of π/3 or 5π/3 on the unit circle, we can write:

2x = π/3 + 2πn or 2x = 5π/3 + 2πn

Solving for x, we divide both sides by 2:

x = π/6 + πn or x = 5π/6 + πn

Here, n is an integer, and each solution corresponds to a value of x for which 2cos(2x) - 1 is equal to zero.

5) tan^2(x) - tan(x) = 0:

To solve this equation, we'll factor out the common term tan(x):

tan(x)(tan(x) - 1) = 0

Now, we have two factors: tan(x) = 0 and tan(x) - 1 = 0.

For tan(x) = 0, we know that tangent is equal to zero at multiples of π. So, we can write:

x = nπ

For tan(x) - 1 = 0, we'll add 1 to both sides:

tan(x) = 1

Using the unit circle or inverse tangent function, we find that:

x = π/4 + πn

Here, n is an integer, and each solution corresponds to a value of x for which tan^2(x) - tan(x) is equal to zero.

These are the solutions to each of the given equations.