Question

1.Решить уравнение: cos3x=sin5x2.Найти все корни уравнения sin2x+16cos^{2}x=4[/tex],принадлежащие отрезку

Answer 1

1. cos3x=sin5x

Или с применением формулы приведения:

sin5x - sin(pi/2 -3x) = 0

Из формулы разности синусов:

2sin[(5x-pi/2 + 3x)/2]*cos[(5x+pi/2 - 3x)/2] = 0

Разбиваем на два уравнения:

sin(4x- pi/4) = 0                 cos(x+ pi/4) = 0

4x- pi/4 = pi*k                    x+ pi/4 = pi/2 + pi*n

x = pi/16 + pi*k/4                x = pi/4 + pi*n

Ответ: pi/16 + pi*k/4 ;   pi/4 + pi*n;   k,n  принадл. Z

 

2. sin2x + 16cos²x = 4

Пользуясь формулой синуса двойного угла и основным тождеством приведем данное уравнение к однородному второй степени:

2sinx*cosx + 16cos²x - 4(sin²x+cos²x)=0

2sin²x - sinx*cosx - 6cos²x = 0

Делим на cos²x:

2tg²x - tgx - 6 = 0,   tgx = t

2t² - t - 6 = 0

D = 1 + 48 = 49 = 7²

t₁ = (1+7)/4 = 2

t₂ = (1-7)/4 = - 1,5

tgx = 2                              tgx = -1,5

x = arctg2 + πk                   x = -arctg1,5 + πn

Подбираем корни из заданного промежутка:

arctg2; π - arctg1,5;  π + arctg2

 

Answer 2

Solving the equation cos(3x) = sin(5x/2):

To solve the equation cos(3x) = sin(5x/2), we can use trigonometric identities and algebraic manipulation. Let's break it down step by step:

Step 1: Convert sin(5x/2) to cos(π/2 - 5x/2) using the identity sin(θ) = cos(π/2 - θ).

Step 2: Rewrite the equation as cos(3x) = cos(π/2 - 5x/2).

Step 3: Apply the identity cos(A) = cos(B) to get two possible cases:

Case 1: 3x = π/2 - 5x/2 Case 2: 3x = -π/2 + 5x/2

Step 4: Solve each case separately:

Case 1: 3x = π/2 - 5x/2 Multiply both sides by 2 to eliminate the fraction: 6x = π - 5x Add 5x to both sides: 11x = π Divide both sides by 11: x = π/11

Case 2: 3x = -π/2 + 5x/2 Multiply both sides by 2 to eliminate the fraction: 6x = -π + 5x Subtract 5x from both sides: x = -π

Therefore, the solutions to the equation cos(3x) = sin(5x/2) are x = π/11 and x = -π.

Finding all the roots of the equation sin(2x) + 16cos^2(x) = 4 on a given interval:

To find all the roots of the equation sin(2x) + 16cos^2(x) = 4 on a given interval, we can use algebraic manipulation and trigonometric identities. Let's break it down step by step:

Step 1: Rewrite the equation as sin(2x) + 16(1 - sin^2(x)) = 4.

Step 2: Simplify the equation: sin(2x) + 16 - 16sin^2(x) = 4 Rearrange the terms: 16sin^2(x) - sin(2x) + 12 = 0

Step 3: Let's solve this quadratic equation for sin(x):

Using the quadratic formula, sin(x) = (-b ± √(b^2 - 4ac)) / (2a), where a = 16, b = -1, and c = 12.

sin(x) = (-(-1) ± √((-1)^2 - 4(16)(12))) / (2(16)) sin(x) = (1 ± √(1 - 768)) / 32 sin(x) = (1 ± √(-767)) / 32

Since the square root of a negative number is not defined in the real number system, there are no real solutions for sin(x) in this case.

Therefore, there are no roots of the equation sin(2x) + 16cos^2(x) = 4 on the given interval.