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Find all solutions of the equation. cosx=- 2

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User Wael
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Cos(x) = -2 has no real solutions as cosine values range from -1 to 1. However, infinite complex solutions exist: x + 2πk, where k is any integer and x = -πi/2 + 2πk. Remember, these complex solutions fall outside the realm of traditional real numbers.

The equation cos(x) = -2 has no real solutions because the cosine function's range is limited to [-1, 1]. This means the output (cos(x)) can never be less than -1 or greater than 1, regardless of the input angle (x). Therefore, achieving a value of -2 is mathematically impossible using the real cosine function.

However, complex solutions exist beyond the realm of real numbers. We can approach this by introducing complex numbers of the form a + bi, where a and b are real numbers and i is the imaginary unit (i^2 = -1).

Using Euler's formula, which connects complex numbers with trigonometric functions, we can express cos(x) as (ei^(ix) + e^(-ix))/2. Setting this equal to -2 and rearranging, we get:

ei^(ix) + e^(-ix) = -4

Factoring out e^(ix), we have:

e^(ix) * (1 + e^(-2ix)) = -4

This equation cannot be solved within the real number system due to the presence of e^(-2ix) that oscillates and never reaches -1. However, if we allow complex solutions, we can find values of x that satisfy the equation.

One possible solution involves taking the square root of both sides, leading to:

e^(ix) = sqrt(-4) = 2i

Since the cosine function has period 2π, its values repeat every 2π radians. Therefore, any solution x + 2πk, where k is an integer, will also be a solution due to the periodicity of cosine.

In conclusion, the equation cos(x) = -2 has no real solutions, but it has infinitely many complex solutions of the form x + 2πk, where k is an integer and x = arccos(2i) = -πi/2 + 2πk.

Remember, these complex solutions lie outside the traditional domain of real numbers and might not be applicable in all contexts.

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User ChristianOConnor
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