Question

2sin(квадрат)x-2cos x=2,5

Answer 1

2sin22x​+5cos2x​=42⋅(1−cos22x​)+5cos2x​=4−2cos22x​+5cos2x​−2=0∣⋅(−1)2cos22x​−5cos2x​+2=0​ 
Пусть \cos\frac{x}{2}=t\,\,\, (|t| \leq 1)cos2x​=t(∣t∣≤1) , получаем
\begin{lgathered}2t^2-5t+2=0\\ D=b^2-4ac=(-5)^2-4\cdot2\cdot2=25-16=9\\ t_1= \frac{5+3}{4} =2\notin[-1;1]\\ t_2= \frac{5-3}{2} = \frac{1}{2}\end{lgathered}2t2−5t+2=0D=b2−4ac=(−5)2−4⋅2⋅2=25−16=9t1​=45+3​=2∉[−1;1]t2​=25−3​=21​​
Обратная замена
\begin{lgathered}\cos\frac{x}{2}=0.5\\ \frac{x}{2}=\pm \frac{\pi}{3} +2 \pi n,n \in \mathbb{Z}\\ x=\pm \frac{2\pi}{3} + 4\pi n,n \in \mathbb{Z}\end{lgathered}cos2x​=0.52x​=±3π​+2πn,n∈Zx=±32π​+4πn,n∈Z​

утверждение может быть неверным

Answer 2

It seems there might be some confusion or a typo in the equation you provided. The variable "квадрат" is not a standard mathematical notation, and it's unclear what it represents. If you meant "квадрат" to be the square of something (like "x^2" in English notation), please clarify.

Assuming you meant to write "2sin^2(x) - 2cos(x) = 2.5," we can attempt to solve the equation.

Let's simplify the equation first:

2sin^2(x) - 2cos(x) = 2.5

Now, we can use the trigonometric identity: sin^2(x) + cos^2(x) = 1 Rearrange this identity: sin^2(x) = 1 - cos^2(x)

Substitute sin^2(x) with 1 - cos^2(x) in the original equation:

2(1 - cos^2(x)) - 2cos(x) = 2.5

Now, distribute the 2 on the left side:

2 - 2cos^2(x) - 2cos(x) = 2.5

Rearrange the equation:

-2cos^2(x) - 2cos(x) + 2 = 2.5

Subtract 2.5 from both sides:

-2cos^2(x) - 2cos(x) - 0.5 = 0

Now, we have a quadratic equation in terms of cos(x):

2cos^2(x) + 2cos(x) + 0.5 = 0

To solve for cos(x), we can use the quadratic formula:

cos(x) = (-b ± √(b^2 - 4ac)) / 2a

In this equation, a = 2, b = 2, and c = 0.5.

cos(x) = (-(2) ± √(2^2 - 4 * 2 * 0.5)) / (2 * 2) cos(x) = (-2 ± √(4 - 4)) / 4 cos(x) = (-2 ± √0) / 4

Since the discriminant (the term inside the square root) is zero, there is only one real solution for cos(x):

cos(x) = (-2 + 0) / 4 cos(x) = -0.5

Now, to find sin(x), we can use the trigonometric identity:

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

sin^2(x) + (-0.5)^2 = 1 sin^2(x) + 0.25 = 1 sin^2(x) = 1 - 0.25 sin^2(x) = 0.75

sin(x) = ±√0.75 ≈ ±0.866

So, the solutions for x are:

1. cos(x) = -0.5 and sin(x) ≈ 0.866
2. cos(x) = -0.5 and sin(x) ≈ -0.866

Keep in mind that this solution assumes you meant "2sin^2(x) - 2cos(x) = 2.5" as the original equation. If "квадрат" represents something different, please provide clarification.