Question

С решением 1. 2sin2x + 9sin x cos x + 4cos2x = 0 2.4 tg x – 6ctg x + 5 = 0 3. 8sin2x + 3sin 2x = 14cos2x 4. 2sin2x – 7cos 2x = 6sin 2x + 7

Answer 1

Ответ:

1) x=-0,5arctg(8/13)+kπ, k∈Ζ

2) x={arctg2+kπ; arctg$\frac{3}{4}$+kπ}, k∈Z

Объяснение:

1)  2sin2x + 9sin x cos x + 4cos2x = 0

9sin x cos x=4,5(2sin x cos x)=4,5sin2x

2sin2x + 4,5sin2x + 4cos2x = 0

6,5sin2x + 4cos2x = 0

Здесь разделим обе части на cos2x≠0. Если cos2x=0

cos2x=0⇒6,5sin2x + 4·0 = 0⇒sin2x=0. Что невозможно, так как должно быть sin²2x+cos²2x=1

6,5tg2x+4=0

tg2x=-8/13

2x=arctg(-8/13)+kπ=-arctg(8/13)+kπ

x=-0,5arctg(8/13)+kπ; k=0;±1;±2;±3;...

2) 4 tg x – 6ctg x + 5 = 0

x≠kπ/2

tgx=t⇒ctgx=1/t

4t-6/t+5=0

4t²+5t-6=0

D=25+96=121

t₁=$\frac{-5-11}{8}=-2$

t₂=$\frac{-5+11}{8}=\frac{6}{8}=\frac{3}{4}$

a) tgx=-2

x=arctg(-2)+kπ=-arctg2+kπ

b) tgx=$\frac{3}{4}$

x=arctg$\frac{3}{4}$+kπ, k∈Z

Answer 2

To solve these trigonometric equations, we'll work on them one by one:

1. 2sin^2(x) + 9sin(x)cos(x) + 4cos^2(x) = 0

To make it easier to work with, let's use the identity sin^2(x) + cos^2(x) = 1:

2(1 - cos^2(x)) + 9sin(x)cos(x) + 4cos^2(x) = 0

Now, we can substitute u = cos(x):

2(1 - u^2) + 9u + 4u^2 = 0

Rearrange and solve the quadratic equation:

4u^2 - 2u - 2 - 9u = 0

4u^2 - 11u - 2 = 0

Using the quadratic formula: u = (-b ± √(b^2 - 4ac)) / 2a, where a = 4, b = -11, and c = -2.

u = (11 ± √(11^2 - 4 * 4 * -2)) / (2 * 4)

u = (11 ± √(121 + 32)) / 8

u = (11 ± √153) / 8

Now, we have two possible values for u:

1. u = (11 + √153) / 8
2. u = (11 - √153) / 8

Next, find the corresponding values for x using the inverse cosine function:

1. x = cos^(-1)((11 + √153) / 8)
2. x = cos^(-1)((11 - √153) / 8)

These are the general solutions to the equation. Remember that there are infinite solutions due to the periodic nature of trigonometric functions. The solutions will depend on the range specified for x.

2. 4tan(x) - 6cot(x) + 5 = 0

Since cot(x) is the reciprocal of tan(x), we can rewrite the equation as:

4tan(x) - 6/tan(x) + 5 = 0

To eliminate the fraction, multiply the whole equation by tan(x):

4tan^2(x) - 6 + 5tan(x) = 0

Now, let y = tan(x):

4y^2 + 5y - 6 = 0

This is a quadratic equation in y. Factor it:

(4y + 6)(y - 1) = 0

Now set each factor equal to zero and solve for y:

1. 4y + 6 = 0 y = -6/4 y = -3/2

2. y - 1 = 0 y = 1

Now, convert these values back to x using the inverse tangent function:

1. x = tan^(-1)(-3/2)

2. x = tan^(-1)(1)

3. 8sin^2(x) + 3sin(2x) = 14cos^2(x)

We'll use the identity sin(2x) = 2sin(x)cos(x):

8sin^2(x) + 6sin(x)cos(x) = 14cos^2(x)

Move all terms to one side:

8sin^2(x) + 6sin(x)cos(x) - 14cos^2(x) = 0

Now, notice that this equation involves both sin and cos terms. Let's use the Pythagorean identity sin^2(x) + cos^2(x) = 1 to replace sin^2(x) and cos^2(x) in the equation:

8(1 - cos^2(x)) + 6sin(x)cos(x) - 14cos^2(x) = 0

Simplify further:

8 - 8cos^2(x) + 6sin(x)cos(x) - 14cos^2(x) = 0

Combine like terms:

-22cos^2(x) + 6sin(x)cos(x) + 8 = 0

Next, factor out cos(x):

cos(x)(-22cos(x) + 6sin(x) + 8) = 0

Now, we have two possible solutions:

1. cos(x) = 0
2. -22cos(x) + 6sin(x) + 8 = 0

For cos(x) = 0, x = π/2 + kπ (where k is an integer) or x = -π/2 + kπ.

For -22cos(x) + 6sin(x) + 8 = 0, we need to solve for x numerically since there is no simple algebraic solution. You can use numerical methods or a graphing calculator to find approximate solutions for x.

4. 2sin^2(x) - 7cos^2(x) = 6sin(2x) + 7

Again, we'll use the identity sin(2x) = 2sin(x)cos(x):

2sin^2(x) - 7cos^2(x) = 6(2sin(x)cos(x)) + 7

Simplify:

2sin^2(x) - 7cos^2(x) = 12sin(x)cos(x) + 7

Now, let's use the Pythagorean identity sin^2(x) + cos^2(x) = 1 to replace sin^2(x) in the equation:

2(1 - cos^2(x)) - 7cos^2(x) = 12sin(x)cos(x) + 7

Expand:

2 - 2cos^2(x) - 7cos^2(x) = 12sin(x)cos(x) + 7

Combine like terms:

-9cos^2(x) = 12sin(x)cos(x) + 5

We can use the identity sin(2x) = 2sin(x)cos(x) to rewrite the right side:

-9cos^2(x) = 6sin(2x) + 5

Rearrange the equation:

6sin(2x) = -9cos^2(x) - 5

Now, use the Pythagorean identity sin^2(x) + cos^2(x) = 1 to replace cos^2(x):

6sin(2x) = -9(1 - sin^2(x)) - 5

6sin(2x) = -9 + 9sin^2(x) - 5

Simplify:

6sin(2x) = 9sin^2(x) - 14

Move all terms to one side:

9sin^2(x) - 6sin(2x) - 14 = 0

Now, we have a quadratic equation in sin(x). Solve for sin(x) and then find the corresponding values of x using the inverse sine function.