Question

1+2sin²x-√3sin2x ------------------------- = 0 2sinx-1

Answer 1

Ответ:

Объяснение:

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Answer 2

To solve the equation (1+2sin²x-√3sin2x)/(2sinx-1) = 0, we need to find the values of x that make the expression equal to zero.

First, let's simplify the equation:

(1 + 2sin²x - √3sin2x) / (2sinx - 1) = 0

To begin, let's factor the numerator:

(1 + sin²x + sin²x - √3sin2x) / (2sinx - 1) = 0

(2sin²x + sin²x - √3sin2x + 1) / (2sinx - 1) = 0

Now, let's rewrite the equation using common terms:

(3sin²x - √3sin2x + 1) / (2sinx - 1) = 0

Next, let's factor the numerator:

((√3sinx - 1)(√3sinx - 1)) / (2sinx - 1) = 0

Now, we can set each factor equal to zero and solve for x:

√3sinx - 1 = 0 or 2sinx - 1 = 0

√3sinx = 1 or 2sinx = 1

sinx = 1/√3 or sinx = 1/2

To find the solutions for x, we need to consider the values of sinx that satisfy these equations.

For sinx = 1/√3, we have two possible solutions:

x₁ = π/3 + 2πn and x₂ = 2π/3 + 2πn

where n is an integer.

For sinx = 1/2, we have two possible solutions:

x₃ = π/6 + 2πn and x₄ = 5π/6 + 2πn

where n is an integer.

Therefore, the solutions to the equation are:

x = π/3 + 2πn, 2π/3 + 2πn, π/6 + 2πn, and 5π/6 + 2πn, where n is an integer.