Question

Tg(пи/5 + 3x)=0 2cos(пи/4 + 2x)= - корень из 2 4sin x/3 * cos x/3 = 2 корень из 3 2cos^2x - 7cosx - 4 = 0

Answer 1

Ответ:

-π/15+πk/3,     -π/8 ±3π/8 + πk, k∈Z,   (-1)^kπ/2 +3πk/2,    ±2π/3 +2πk, k∈Z

Пошаговое объяснение:

1) tg(π/5+3x) = 0

π/5+3x=arctg0+πk, k∈Z

π/5+3x= πk

3x= -π/5 +πk

x= -π/15+πk/3, k∈Z

2) cos(π/4+2x)= -√2/2

π/4+2x=±arccos(-√2/2)+2πk, k∈Z

2x=-π/4 ±3π/4+2πk

x= - π/8 ±3π/8+πk, k∈Z

3) 4sin(x/3)cos(x/3) = 2√3

2sin(2x/3) = √3

sin(2x/3)=√3/2

2x/3=( -1)^k arcsin(√3/2)+πk

2x/3 = (-1)^k π/3 + πk

x= ( -1)^k π/2 + 3πk/2, k∈Z

4) D=(-7)²- 4*2*(-4)=81

cosx=(7±9)/4

cosx =4, решений нет;

cosx= -1/2

x= ±arccos(-1/2) +2πn

x= ± 2π/3+2πn, n∈Z

Answer 2

To solve each of the given trigonometric equations, we'll go through them one by one:

1. Tg(π/5 + 3x) = 0:

The tangent function is equal to zero when its argument is an integer multiple of π. Therefore, we can write:

π/5 + 3x = nπ, where n is an integer.

Now, solve for x:

3x = nπ - π/5 x = (nπ - π/5) / 3

2. 2cos(π/4 + 2x) = -√2:

Divide both sides by 2:

cos(π/4 + 2x) = -√2 / 2

The cosine function is equal to -√2/2 at certain special angles, such as π/3 and 5π/3:

π/4 + 2x = π/3 + 2nπ or π/4 + 2x = 5π/3 + 2nπ, where n is an integer.

Now, solve for x in each case:

a) π/4 + 2x = π/3 + 2nπ 2x = π/3 - π/4 + 2nπ x = (π/3 - π/4 + 2nπ) / 2

b) π/4 + 2x = 5π/3 + 2nπ 2x = 5π/3 - π/4 + 2nπ x = (5π/3 - π/4 + 2nπ) / 2

3. 4sin(x/3) * cos(x/3) = 2√3:

Use the double-angle identity for sine: sin(2θ) = 2sin(θ)cos(θ).

4sin(x/3) * cos(x/3) = 2sin(2x/3) = 2√3

Now, solve for 2x/3:

2x/3 = π/3 + 2nπ or 2x/3 = 2π/3 + 2nπ, where n is an integer.

Now, solve for x in each case:

a) 2x/3 = π/3 + 2nπ x = (π/3 + 2nπ) * 3/2

b) 2x/3 = 2π/3 + 2nπ x = (2π/3 + 2nπ) * 3/2

4. 2cos^2(x) - 7cos(x) - 4 = 0:

Let cos(x) = t, then the equation becomes:

2t^2 - 7t - 4 = 0

Now, solve this quadratic equation for t:

Using the quadratic formula: t = (-b ± √(b^2 - 4ac)) / 2a

where a = 2, b = -7, and c = -4

t = [7 ± √((-7)^2 - 4 * 2 * (-4))] / 2 * 2 t = [7 ± √(49 + 32)] / 4 t = [7 ± √81] / 4 t = [7 ± 9] / 4

Now, we have two possible values for t:

a) t = (7 + 9) / 4 = 16 / 4 = 4 b) t = (7 - 9) / 4 = -2 / 4 = -1/2

Recall that t = cos(x). Now, find the corresponding values of x for each case:

a) cos(x) = 4 Since the cosine function has a range of [-1, 1], there are no real solutions for this case.

b) cos(x) = -1/2 x = π ± 2nπ or x = 2π ± 2nπ, where n is an integer.

Now, we have found the solutions to each of the given trigonometric equations. Remember to check the domain of the original equations and verify if the solutions fall within the specified domain.