2 sin(пи+x)*cos(пи/2+x)=sin x [-5пи; -4 пи]

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вопрос

Let's simplify the expression and calculate the given trigonometric equation:

2 sin(π + x) * cos(π/2 + x) = sin x

Using trigonometric identities, we know that sin(π + x) = -sin(x), and cos(π/2 + x) = -sin(x). Substituting these values into the equation:

2(-sin(x))(-sin(x)) = sin(x)
2sin(x)*sin(x) = sin(x)
2sin^2(x) = sin(x)
2sin^2(x) - sin(x) = 0
sin(x)(2sin(x) - 1) = 0

Therefore, the solutions for sin(x) = 0 and 2sin(x) - 1 = 0:

For sin(x) = 0:
x = -5π, -4π (within the interval given)

For 2sin(x) - 1 = 0:
sin(x) = 1/2
x = π/6, 5π/6

The solutions within the interval [-5π, -4π] are x = -5π, -4π.