Question

Sin(2x+3п)sin(3x+3п/2)=0

Answer 1

Ответ:

$\left \{ {{ x=\frac{\pi k}{2}} \atop {x = \frac{\pi}{6} +\frac{\pi k}{3}}} \right. \, \, \, , k \in Z$

Объяснение:

$sin(2x+3\pi)=-sin2x\\sin(3x+\frac{3\pi}{2})=-cos3x \\\\sin(2x+3\pi)sin(3x+\frac{3\pi}{2})=0\\\\sin2xcos3x=0$

1:

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Answer 2

To solve the trigonometric equation sin(2x + 3π)sin(3x + 3π/2) = 0, we need to find the values of x that satisfy this equation. The product of two factors is equal to zero if and only if at least one of the factors is equal to zero.

So, we have two cases to consider:

Case 1: sin(2x + 3π) = 0 To find the values of x for this case, we set the first factor equal to zero and solve for x:

2x + 3π = kπ, where k is an integer. 2x = kπ - 3π 2x = (k - 3)π x = (k - 3)π/2, where k is an integer.

Case 2: sin(3x + 3π/2) = 0 To find the values of x for this case, we set the second factor equal to zero and solve for x:

3x + 3π/2 = mπ, where m is an integer. 3x = mπ - 3π/2 3x = (2m - 3)π/2 x = (2m - 3)π/6, where m is an integer.

Now, combine the solutions from both cases:

1. For Case 1, x = (k - 3)π/2, where k is an integer.
2. For Case 2, x = (2m - 3)π/6, where m is an integer.

These are the general solutions to the equation sin(2x + 3π)sin(3x + 3π/2) = 0. You can also simplify and combine them further if needed.