2 cos 2(x-pi/3)+8cos(x-pi/3)=3

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Let's start by simplifying the equation:

2cos(2(x-pi/3)) + 8cos(x-pi/3) = 3

Using the double angle identity for cosine: cos(2x) = 2cos(x)^2 - 1, we can simplify the expression for 2cos(2(x-pi/3)):

2cos(2(x-pi/3)) = 2[2cos(x-pi/3)^2 - 1]

Now we will simplify the equation further:

2[2cos(x-pi/3)^2 - 1] + 8cos(x-pi/3) = 3
4cos(x-pi/3)^2 - 2 + 8cos(x-pi/3) = 3
4cos(x-pi/3)^2 + 8cos(x-pi/3) - 5 = 0

Let u = cos(x-pi/3), then the equation becomes:

4u^2 + 8u - 5 = 0

Now we can solve this quadratic equation for u using the quadratic formula:

u = (-b ± √(b^2 - 4ac)) / 2a

Plugging in the values of a, b, and c into the formula:

u = (-8 ± √(8^2 - 44(-5))) / (2*4)
u = (-8 ± √(64 + 80)) / 8
u = (-8 ± √144) / 8
u = (-8 ± 12) / 8

This gives us two possible values for u:

u = (-8 + 12) / 8 = 4 / 8 = 0.5
u = (-8 - 12) / 8 = -20 / 8 = -2.5

Now we need to convert these values of u back to cos(x - pi/3) by substituting back:

cos(x - pi/3) = 0.5
x - pi/3 = arccos(0.5)
x - pi/3 = pi/3 or x - pi/3 = 5pi/3
x = 2pi/3 or x = 2pi

cos(x - pi/3) = -2.5
This would not be possible as the cosine function can only have values from -1 to 1.

Therefore, the solutions to the equation are x = 2pi/3 and x = 2pi.