2cos^2x+2√2cos(pi/2-x)+1=0 Б)[3pi/2;3pi]

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To solve this equation, we will first rewrite it in terms of cos(x) and then solve for cos(x):

2cos^2(x) + 2√2cos(π/2 - x) + 1 = 0
2cos^2(x) + 2√2sin(x) + 1 = 0

Now, let's use the double angle formula for cosine to express cos^2(x) in terms of sine:

cos(2x) = 2cos^2(x) - 1
cos^2(x) = (cos(2x) + 1)/2

Substitute this back into the equation:

2[(cos(2x) + 1)/2] + 2√2sin(x) + 1 = 0
cos(2x) + 1 + 2√2sin(x) + 1 = 0
cos(2x) + 2√2sin(x) + 2 = 0

Now, let's use the angle addition formula for cosine to express cos(2x) in terms of cos(x) and sin(x):

cos(2x) = cos^2(x) - sin^2(x)
cos(2x) = cos^2(x) - (1 - cos^2(x))
cos(2x) = 2cos^2(x) - 1

Substitute this back into the equation:

2cos^2(x) - 1 + 2√2sin(x) + 2 = 0
2cos^2(x) + 2√2sin(x) + 1 = 0

This is the same equation we had before, so we have:

2cos^2(x) + 2√2sin(x) + 1 = 0

Since sin(x) = cos(π/2 - x), we can rewrite the equation as:

2cos^2(x) + 2√2cos(π/2 - x) + 1 = 0

Therefore, the given equation is equivalent to 2cos^2(x) + 2√2cos(π/2 - x) + 1 = 0.

To solve for x in the interval [3π/2, 3π], we can apply the quadratic formula with a = 2, b = 2√2, and c = 1 to find the values of cos(x) that satisfy the equation. Once we find cos(x), we can then find x by taking the inverse cosine.

Let's solve the equation using the quadratic formula:

cos(x) = [-b ± √(b^2 - 4ac)] / 2a
cos(x) = [-2√2 ± √(8 - 8)] / 4
cos(x) = [-2√2 ± 0] / 4
cos(x) = ±√2 / 2

Since cos(x) can only take values between -1 and 1, we have:

cos(x) = √2 / 2

Now we find x by taking the inverse cosine:

x = cos^(-1)(√2 / 2)
x = π/4

Therefore, the solution to the equation 2cos^2(x) + 2√2cos(π/2 - x) + 1 = 0 in the interval [3π/2, 3π] is x = π/4.