Question

Cos4x+кореньиз2cos2x+1=0

Answer 1

$\cos4x+\sqrt{2} \cos 2x+1=0\\ 2\cos^22x-1+\sqrt{2} \cos2x+1=0\\ 2\cos^22x+\sqrt{2} \cos 2x=0\\ \cos2x(2\cos2x+\sqrt{2} )=0$

Произведение равно нулю, если хотя бы один из множителей равен нулю.

$\displaystyle \left[\begin{array}{ccc}\cos2x=0\\ \\ \cos2x=-\frac{\sqrt{2} }{2}\end{array}\right ~~~\Rightarrow~~\left[\begin{array}{ccc}2x=\frac{\pi}{2}+\pi n,n \in \mathbb{Z}\\ \\ 2x=\pm\frac{3\pi}{4}+2\pi n,n \in \mathbb{Z}\end{array}\right ~~\Rightarrow~~\\ \\ \\ \Rightarrow~~\left[\begin{array}{ccc}x=\frac{\pi}{4}+\frac{\pi n}{2}, n\in \mathbb{Z}\\ \\ x=\pm\frac{3\pi}{8}+\pi n,n \in \mathbb{Z}\end{array}\right$

Answer 2

To solve the equation cos(4x) + √(2cos(2x) + 1) = 0, we'll first simplify it and then find the solutions.

Let's start with the expression inside the square root:

√(2cos(2x) + 1)

Now, we know that 2cos(2x) can be written as 2 * cos^2(x) - 1 (using the double-angle formula for cosine).

So, the expression becomes:

√(2 * cos^2(x) - 1 + 1)

Simplifying further:

√(2 * cos^2(x))

Taking the square root of 2:

√2 * √(cos^2(x))

Since cosine is always positive or zero, we can drop the absolute value signs:

√2 * cos(x)

Now, our original equation becomes:

cos(4x) + √2 * cos(x) = 0

Now, we have a sum of two trigonometric functions. Unfortunately, this equation cannot be solved algebraically in a straightforward manner. However, we can use numerical methods or graphical methods to approximate the solutions.

One common numerical method is to use iterative techniques like Newton's method. But since you didn't specify any range for the solutions, I'll provide you with a graph to visualize the solutions.

Let's graph the left-hand side of the equation, y = cos(4x) + √2 * cos(x), and find the x-values (angles) where it intersects with the x-axis (y = 0).

Here's the graph: (Please note that I can't generate a live graph here, so I'll describe the general shape.)

• The graph will show a series of peaks and troughs because of the oscillating nature of the cosine function.
• The graph will intersect the x-axis at various points. These x-values are the approximate solutions to the equation.

By visually examining the graph, you can estimate the values of x where y is close to zero. If you need more precise solutions, you can use numerical methods as mentioned earlier.

Keep in mind that trigonometric equations often have infinitely many solutions, so you may need to specify a particular interval or domain for the solutions you are interested in.