Question

Arccos(x^2+3x-2)=arccos(3-x^2)​

Answer 1

Ответ:

Объяснение:

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Answer 2

To solve the equation arccos(x^2 + 3x - 2) = arccos(3 - x^2), we can start by recognizing that the arccos function has a range of [0, π] (or [0°, 180°] in degrees). This means that the values of arccos(x^2 + 3x - 2) and arccos(3 - x^2) must lie within this range.

Since the arccos function is a one-to-one function, we can equate the arguments inside the arccos functions:

x^2 + 3x - 2 = 3 - x^2

Rearranging the equation, we have:

2x^2 + 3x - 5 = 0

Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. In this case, we'll use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 2, b = 3, and c = -5. Plugging these values into the formula:

x = (-3 ± √(3^2 - 4 * 2 * -5)) / (2 * 2)

Simplifying further:

x = (-3 ± √(9 + 40)) / 4 x = (-3 ± √49) / 4 x = (-3 ± 7) / 4

This gives us two possible solutions:

1. x = (-3 + 7) / 4 = 1
2. x = (-3 - 7) / 4 = -2.5

Therefore, the equation arccos(x^2 + 3x - 2) = arccos(3 - x^2) has two solutions: x = 1 and x = -2.5.