Question

2cos^2 3x+5sin 3x-4=0

Answer 1

Из основного тригонометрического тождества: cos^2(3x) = 1 - sin^2(3x). Тогда исходное уравнение примет вид:
2 - 2sin^2(3x) + 5sin(3x) - 4 = 0;
2sin^2(3x) - 5sin(3x) + 2 = 0;
Пусть sin(3x) = t. Тогда:
2t^2 - 5t + 2 = 0;
D = 25 - 4*2*2 = 9.
t = (5-3)/4 = 1/2;
ИЛИ
t = (5+3)/4 = 2.
Вернемся к синусу:
sin(3x) = 2. Это уравнение не имеет корней, так как область значений функции sin(t) - это промежуток [-1; 1].
sin(3x) = 1/2;
3x = (-1)^k * π/6 + πk, k∈Z;
x = (-1)^k * π/18 + πk/3, k∈Z.

Ответ: x = (-1)^k * π/18 + πk/3, k∈Z.

Answer 2

To solve the trigonometric equation 2cos^2(3x) + 5sin(3x) - 4 = 0, we need to find the values of x that satisfy the equation.

First, let's make use of the trigonometric identity: cos^2θ + sin^2θ = 1. Rearranging the equation:

2cos^2(3x) + 5sin(3x) - 4 = 0

2cos^2(3x) - 4 = -5sin(3x)

Divide both sides by 2:

cos^2(3x) - 2 = -2.5sin(3x)

Now, we can use the Pythagorean identity sin^2θ + cos^2θ = 1 to replace cos^2(3x):

1 - sin^2(3x) - 2 = -2.5sin(3x)

Simplify:

1 - sin^2(3x) = -2.5sin(3x) + 2

Rearrange to a quadratic equation:

sin^2(3x) + 2.5sin(3x) - 3 = 0

Now, we can solve this quadratic equation for sin(3x) and then find the values of x from there.

Let's introduce a substitution: Let y = sin(3x)

The equation now becomes:

y^2 + 2.5y - 3 = 0

To solve this quadratic equation, we can factor it:

(y + 3)(y - 1) = 0

Now set each factor to zero and solve for y:

y + 3 = 0 or y - 1 = 0

If y + 3 = 0, then y = -3

If y - 1 = 0, then y = 1

Now, recall the substitution y = sin(3x):

1. sin(3x) = -3
2. sin(3x) = 1

However, there is no real value for sin(3x) that equals -3, as the sine function is bound between -1 and 1. Thus, the first equation has no real solution.

Now, let's solve the second equation:

sin(3x) = 1

To find the solutions for this, we need to look at the values of sine for which it equals 1. The sine function reaches its maximum value of 1 at π/2 (90 degrees) and any integer multiple of 2π.

So, 3x = π/2 + 2kπ, where k is an integer.

Solving for x:

x = (π/2 + 2kπ) / 3, where k is an integer.

These are the solutions to the equation 2cos^2(3x) + 5sin(3x) - 4 = 0. Keep in mind that there may be other solutions beyond the given range depending on the domain of the problem.