Sin 15 + sin 3x/2 = 0

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вопрос

To solve the equation sin(15) + sin(3x / 2) = 0, we need to find the values of x that satisfy the equation.

First, we can rewrite sin(15) as sin(3π / 12) and sin(3x / 2) as sin(πx / 12). Using the sum-to-product formula for sine, we have:

sin(a) + sin(b) = 2sin((a + b) / 2)cos((a - b) / 2)

Therefore, the given equation can be rewritten as:

2sin((3π / 12 + πx / 12) / 2)cos((3π / 12 - πx / 12) / 2) = 0

Simplifying, we get:

2sin((4π + πx) / 24)cos((2π - πx) / 24) = 0

This implies sin((4π + πx) / 24) = 0 or cos((2π - πx) / 24) = 0.

For sin((4π + πx) / 24) = 0:

(4π + πx) / 24 = (2n + 1)π, where n is an intege
4π + πx = 24(2n + 1
πx = 48n + 2
x = 48n / π + 20 / π

For cos((2π - πx) / 24) = 0:

(2π - πx) / 24 = (2m + 1)π / 2, where m is an intege
2π - πx = 12(2m + 1
πx = 24m +
x = 24m / π + 4 / π

Therefore, the solutions for x are x = 48n / π + 20 / π or x = 24m / π + 4 / π, where n and m are integers.