The equation cos(x) + cos(2x) + cos(3x) = 1 has no real solutions.
To see this, we can use the identity cos(a) + cos(b) = 2*cos((a+b)/2)*cos((a-b)/2) to rewrite the given equation as follows:
cos(x) + cos(2x) + cos(3x) = 1
2*cos(3x/2)*cos(x/2) = 1
Since the cosine function has a maximum value of 1, the left-hand side of this equation can never be equal to 1. Therefore, there are no real solutions to the equation cos(x) + cos(2x) + cos(3x) = 1.
Here is a proof by contradiction:
Assume that there exists a real number x such that cos(x) + cos(2x) + cos(3x) = 1. Then,
2*cos(3x/2)*cos(x/2) = 1
Since the cosine function has a maximum value of 1, this implies that cos(3x/2) = 1 and cos(x/2) = 1/2. But this is a contradiction, because cos(x/2) can only be 1 if x/2 = 0 or x/2 = pi, and neither of these values satisfy the equation cos(3x/2) = 1.
Therefore, our assumption must be false, and there are no real solutions to the equation cos(x) + cos(2x) + cos(3x) = 1.
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Answer:
The equation cos(x) + cos(2x) + cos(3x) = 1 has no real solutions.
To see this, we can use the identity cos(a) + cos(b) = 2*cos((a+b)/2)*cos((a-b)/2) to rewrite the given equation as follows:
cos(x) + cos(2x) + cos(3x) = 1
2*cos(3x/2)*cos(x/2) = 1
Since the cosine function has a maximum value of 1, the left-hand side of this equation can never be equal to 1. Therefore, there are no real solutions to the equation cos(x) + cos(2x) + cos(3x) = 1.
Here is a proof by contradiction:
Assume that there exists a real number x such that cos(x) + cos(2x) + cos(3x) = 1. Then,
2*cos(3x/2)*cos(x/2) = 1
Since the cosine function has a maximum value of 1, this implies that cos(3x/2) = 1 and cos(x/2) = 1/2. But this is a contradiction, because cos(x/2) can only be 1 if x/2 = 0 or x/2 = pi, and neither of these values satisfy the equation cos(3x/2) = 1.
Therefore, our assumption must be false, and there are no real solutions to the equation cos(x) + cos(2x) + cos(3x) = 1.
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Step-by-step explanation:
The given equation is:
cos(x) + cos(2x) + cos(3x) = 1
Let's use the identity cos(a) + cos(b) = 2cos((a+b)/2)cos((a-b)/2) to simplify the equation:
cos(2x) + cos(x + 2x) = 2cos(2.5x)cos(0.5x)
cos(2x) + cos(3x) = 2cos(2.5x)cos(0.5x)
Now, let's use the identity cos(a) - cos(b) = -2sin((a+b)/2)sin((a-b)/2) to simplify the equation further:
2sin(2.5x)sin(0.5x) = cos(2x) - cos(3x)
2sin(2.5x)sin(0.5x) = -2sin(2x)sin(x)
sin(2.5x)sin(0.5x) = -sin(2x)sin(x)
Now, let's use the identity sin(a)sin(b) = 0.5(cos(a-b) - cos(a+b)) to simplify the equation even further:
0.5(cos(2x - 2.5x) - cos(2x + 2.5x)) = -0.5(cos(x - 2x) - cos(x + 2x))
0.5(cos(-0.5x) - cos(4.5x)) = 0.5(cos(x) - cos(3x))
cos(4.5x) - cos(0.5x) = cos(3x) - cos(x)
2sin(2.5x)sin(1.5x) = -2sin(2x)sin(x)
sin(2.5x)sin(1.5x) = -sin(2x)sin(x)
Now, we have two possible solutions:
1. sin(2.5x) = 0 or sin(1.5x) = 0
2. sin(2x) = -sin(0.5x)
If we solve for the first solution, we get:
x = 0, pi/2, pi, 3pi/2, 2pi/3, 4pi/3
If we solve for the second solution