Cos(2x)+cos(4x)+cos(6x)=0

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вопрос

To solve the equation cos(2x) + cos(4x) + cos(6x) = 0, we can use the trigonometric identity formula:

cos(a) + cos(b) = 2cos((a+b)/2)cos((a-b)/2)

Therefore, we can rewrite the equation as:

2cos(3x)cos(x) = -cos(5x)

Using the trigonometric identity formula again for the cosine of the sum of angles, we have:

cos(a) + cos(b) = 2cos((a+b)/2)cos((a-b)/2)

cos(3x) + cos(2x) = 2cos((3x+2x)/2)cos((3x-2x)/2)
cos(3x) + cos(2x) = 2cos(5x/2)cos(x)

Substitute this into the original equation:

2cos(3x)cos(x) = -2cos(5x/2)cos(x)

Thus, cos(3x) = -cos(5x/2)

Therefore, the solutions for the original equation cos(2x) + cos(4x) + cos(6x) = 0 are any values of x that satisfy the relationship cos(3x) = -cos(5x/2).