Question

Решите 12cos²x+10sinx=4

Answer 1

12cos²x + 10sinx = 4   ║: 2

6cos²x + 5sinx - 2 = 0

6•(1 - sin²x) + 5sinx - 2 = 0

6 - 6sin²x + 5sinx - 2 = 0

- 6sin²x + 5sinx + 4 = 0

6sin²x - 5sinx - 4 = 0

Пусть sinx = a , a ∈ [ - 1 ; 1 ] , тогда

6a² - 5a - 4 = 0

D = (-5)² - 4•6•(-4) = 25 + 96 = 121 = 11²

a₁ = (5 - 11)/12 = - 6/12 = - 1/2

a₂ = (5 + 11)/12 = 16/12 = 4/3 ∉ [ - 1 ; 1 ]

a = - 1/2  ⇔  sinx = - 1/2

[ x = - (п/6) + 2пn

[ x = - (5п/6) + 2пn , n ∈ Z

ОТВЕТ: - (п/6) + 2пn ; - (5п/6) + 2пn , n ∈ Z

Answer 2

To solve the equation 12cos²x + 10sinx = 4, we can use some trigonometric identities to simplify it. First, recall the identity:

cos²x + sin²x = 1

Now, let's rewrite the given equation:

12cos²x + 10sinx = 4

Divide the entire equation by 4 to make it easier to work with:

3cos²x + (5/2)sinx = 1

Now, let's use the identity cos²x + sin²x = 1 to replace cos²x with (1 - sin²x):

3(1 - sin²x) + (5/2)sinx = 1

Expand the equation:

3 - 3sin²x + (5/2)sinx = 1

Move all terms to one side of the equation:

3sin²x - (5/2)sinx + 2 = 0

This is now a quadratic equation in terms of sinx. To solve it, we can use the quadratic formula:

For an equation of the form ax² + bx + c = 0, the solutions for x are given by:

x = (-b ± √(b² - 4ac)) / 2a

In our case, a = 3, b = -(5/2), and c = 2. Now, plug these values into the quadratic formula:

sinx = [-(5/2) ± √((5/2)² - 4 * 3 * 2)] / 2 * 3 sinx = [-(5/2) ± √(25/4 - 24)] / 6 sinx = [-(5/2) ± √(25/4 - 96/4)] / 6 sinx = [-(5/2) ± √(-71/4)] / 6

Since the value inside the square root is negative, the equation has no real solutions. Therefore, the original equation 12cos²x + 10sinx = 4 has no real solutions as well.