Question

Sin^4x+cos^4x+cos2x=0,5

Answer 1

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$\displaystyle cos2x+2=0\\\\cos2x=-2$

решений нет

Ответ х=π/4+πn/2; n∈Z

Answer 2

We can start by using the identity cos^2x + sin^2x = 1, which implies that cos^4x + sin^4x = (cos^2x + sin^2x)^2 - 2cos^2x sin^2x = 1 - 2cos^2x sin^2x.

Then, we can rewrite the equation as:

1 - 2cos^2x sin^2x + cos2x = 0.5

Multiplying both sides by 2, we get:

2 - 4cos^2x sin^2x + 2cos2x = 1

Now, we can use the double angle formula for cosine, which states that cos2x = 2cos^2x - 1, to substitute cos2x in the equation:

2 - 4cos^2x sin^2x + 4cos^2x - 2 = 1

Simplifying the expression, we get:

4cos^2x(1 - sin^2x) - 1 = 0

Using the identity sin^2x = 1 - cos^2x, we can substitute sin^2x in terms of cos^2x:

4cos^2x(cos^2x) - 4cos^4x - 1 = 0

Multiplying by -1 and rearranging, we obtain a quadratic equation:

4cos^4x - 4cos^2x - 1 = 0

We can solve for cos^2x using the quadratic formula:

cos^2x = [4 ± √(16 + 16)]/8 = [1 ± √2]/2

Since cosine is a positive even function, we only need to consider the positive root:

cos^2x = [1 + √2]/2

Taking the square root, we get:

cosx = ±√[(1 + √2)/2]

Finally, we can use the identity sin^2x = 1 - cos^2x to obtain the values of sinx:

sinx = ±√[(1 - √2)/2] or sinx = ±√[(√2 - 1)/2]

Therefore, the solutions to the equation are:

x = nπ ± arcsin[√(1 - (1 + √2)/2)] or x = nπ ± arcsin[√((√2 - 1)/2)], where n is an integer.