Question

Cos(x) + 2cos(2x+π/6)=√3cos(2x)

Answer 1

Ответ:

решение представлено на фото

Answer 2

$\cos x+2\cos(2x+\frac{\pi}{6})=\sqrt{3}\cos 2x\\ \\ \cos x+2(\cos 2x\cos \frac{\pi}{6}-\sin 2x\sin \frac{\pi}{6})=\sqrt{3}\cos 2x\\ \\ \cos x+2(\frac{\sqrt{3}}{2}\cos 2x-\sin 2x\cdot \frac{1}{2})=\sqrt{3}\cos 2x\\ \\ \cos x+\sqrt{3}\cos 2x-\sin 2x=\sqrt{3}\cos 2x\\ \\ \cos x-2\sin x\cos x=0\\ \\ \cos x(1-2\sin x)=0$

Произведение равно нулю в том случае, когда хотя бы один из множителей обращается к нулю.

$\cos x=0~~~\Rightarrow~~~ \boxed{x_1=\frac{\pi}{2}+\pi n,n \in \mathbb{Z}}\\ \\ 1-2\sin x=0\\ \\ \sin x=\frac{1}{2}~~~\Rightarrow~~~ \boxed{x_2=(-1)^k\cdot \frac{\pi}{6}+\pi k,k \in \mathbb{Z}}$

Answer 3

To solve the equation cos(x) + 2cos(2x + π/6) = √3cos(2x), we can start by simplifying it step by step.

Let's begin by expanding the cosine functions:

cos(x) + 2cos(2x)cos(π/6) - 2sin(2x)sin(π/6) = √3cos(2x)

Simplifying further, we know that cos(π/6) = √3/2 and sin(π/6) = 1/2:

cos(x) + 2(2cos²(x) - 1)(√3/2) - 2sin(2x)(1/2) = √3cos(2x)

cos(x) + (√3/2)(4cos²(x) - 2) - sin(2x) = √3cos(2x)

Now, let's simplify the expression by combining like terms:

cos(x) + 2√3cos²(x) - √3 - sin(2x) = √3cos(2x)

Next, we'll rewrite the sine function as a cosine function using the identity sin(2x) = 2sin(x)cos(x):

cos(x) + 2√3cos²(x) - √3 - 2sin(x)cos(x) = √3cos(2x)

Now, we can rearrange the terms:

cos(x) - 2sin(x)cos(x) + 2√3cos²(x) - √3 = √3cos(2x)

We can further simplify the left side by factoring out a common factor of cos(x):

cos(x)(1 - 2sin(x) + 2√3cos(x)) - √3 = √3cos(2x)

Now, we have a common factor of cos(x) on both sides of the equation:

cos(x)(1 - 2sin(x) + 2√3cos(x) - √3cos(2x)) = √3

To solve this equation, we need to consider multiple cases where cos(x) = 0 or cos(x) ≠ 0.

Case 1: cos(x) = 0 If cos(x) = 0, then the equation becomes:

0(1 - 2sin(x) + 2√3cos(x) - √3cos(2x)) = √3

This simplifies to 0 = √3, which is not possible. Therefore, there are no solutions in this case.

Case 2: cos(x) ≠ 0 If cos(x) ≠ 0, we can divide both sides of the equation by cos(x):

1 - 2sin(x) + 2√3cos(x) - √3cos(2x) = √3/cos(x)

Simplifying the right side, we have:

1 - 2sin(x) + 2√3cos(x) - √3cos(2x) = √3sec(x)

From here, you can continue solving the equation by using trigonometric identities, factoring, or applying other techniques depending on the specific problem or restrictions given.