Cos квадрат x+ 4 cos x+3=0

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This equation can be rewritten as:

cos^2(x) + 4cos(x) + 3 = 0

Now, let's substitute cos(x) = y:

y^2 + 4y + 3 = 0

This is a quadratic equation that can be factored as:

(y + 3)(y + 1) = 0

So, the solutions to this equation are y = -3 and y = -1.

Now substitute back cos(x) for y:

cos(x) = -3 or cos(x) = -1

However, the range of the cosine function is -1 ≤ cos(x) ≤ 1. Therefore, cos(x) = -3 is not a valid solution.

Therefore, the only valid solution is cos(x) = -1.

This means that x = π or x = 2π.

So, the solutions are x = π and x = 2π.