Answer:
= (6π - πi)/6
Explanation:
We know that the range of arccosine is [0, π]. Since -√3/2 is less than -1, there is no real number x such that cos(x) = -√3/2. Therefore, the equation arccos(-√3/2) has no solution in the real numbers.
However, if we extend the range of the arccosine function to include complex numbers, we can find a solution using the formula:
arccos(z) = π - i ln(z + i√(1-z²))
where ln denotes the natural logarithm.
Substituting z = -√3/2, we get:
arccos(-√3/2) = π - i ln(-√3/2 + i√(1-(-√3/2)²))
Simplifying the expression under the natural logarithm, we get:
arccos(-√3/2) = π - i ln(-√3/2 + i/2)
To evaluate the natural logarithm, we can write -√3/2 + i/2 in polar form:
-√3/2 + i/2 = e^(i(π/6))
Therefore, we have:
arccos(-√3/2) = π - i ln(e^(i(π/6)))
Using the property ln(e^z) = z of the natural logarithm, we can simplify further:
arccos(-√3/2) = π - i (π/6)
Simplifying the expression, we get:
arccos(-√3/2) = (6π - πi)/6
Therefore, the exact value of arccos(-√3/2) in radians is (6π - πi)/6.
I hope this help