2sin(3п/2-x)*sin(x-п)+(корень из 2)cosx=0

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вопрос

To solve the given trigonometric equation:

2sin(3π/2 - x) sin(x - π) + √2 cos(x) = 0

We can use the trigonometric identity sin(a - b) = sin(a)cos(b) - cos(a)sin(b) to simplify the equation.

Therefore, the equation becomes:

2sin(3π/2 - x) sin(x - π) + √2 cos(x) = 2(sin(3π/2)cos(x) - cos(3π/2)sin(x)) (sin(x)cos(π) - cos(x)sin(π)) + √2 cos(x)
= 2(-1 cos(x)) (0 - 1 sin(x)) + √2 cos(x)
= 2cos(x)sin(x) + √2 * cos(x)

Factor out cos(x) from the equation:

cos(x) * (2sin(x) + √2) = 0

Now, we have two cases to consider:

cos(x) = 0
This occurs when x = π/2 + πn for any integer n.

2sin(x) + √2 = 0
sin(x) = -√2/2
This occurs when x = 3π/4 + 2πn or x = 5π/4 + 2πn for any integer n.

Therefore, the solutions to the trigonometric equation are x = π/2 + πn, x = 3π/4 + 2πn, and x = 5π/4 + 2πn for any integer n.