Question

DUN! Uud NPL, cos²x+ 2 sinx+2=0

Answer 1

Ответ: x=3π/2+2πn.

Пошаговое объяснение:

cos²x+2*sinx+2=0

1-sin²x+2*sinx+2=0

sin²x-2*sinx-3=0

Пусть sinx=t   ⇒

t²-2t-3=0    D=16     √D=4

t₁=sinx=3  ∉ тка как |sinx|≤1

t₂=sinx=-1      x=3π/2+2πn.

Answer 2

To solve the equation cos²x + 2sinx + 2 = 0, we can use some trigonometric identities and algebraic manipulation. Let's go step by step:

1. Start by using the identity cos²x = 1 - sin²x to rewrite the equation: (1 - sin²x) + 2sinx + 2 = 0

2. Rearrange the terms: -sin²x + 2sinx + 3 = 0

3. Multiply the entire equation by -1 to change the signs: sin²x - 2sinx - 3 = 0

4. Now, let's introduce a substitution. Let's say y = sinx. We can rewrite the equation in terms of y: y² - 2y - 3 = 0

5. Factorize the quadratic equation: (y - 3)(y + 1) = 0

6. Solve for y by setting each factor to zero: y - 3 = 0 or y + 1 = 0

   Solving these equations: y = 3 or y = -1

7. Now, substitute back sinx for y in each case: Case 1: sinx = 3 Since the range of sine function is [-1, 1], sinx cannot be equal to 3. So, this case is not valid.

   Case 2: sinx = -1 Solving this equation, we get: x = -π/2 + 2πn (where n is an integer)

   Therefore, the solution to the equation cos²x + 2sinx + 2 = 0 is: x = -π/2 + 2πn (where n is an integer)