Answer
### a) Решение уравнения
Уравнение:
\[
\cos \left(\frac{3 \pi}{2} + 2x\right) = \cos x
\]
Корни:
\[
x = \frac{\pi}{6} + 2k\pi, \quad x = \frac{\pi}{2} + k\pi, \quad x = \frac{5\pi}{6} + 2k\pi, \quad k \in \mathbb{Z}
\]
### б) Корни на отрезке \(\left[\frac{5\pi}{2}; 4\pi\right]\)
Корни:
\[
x = \frac{\pi}{2} + 2\pi, \quad x = \frac{5\pi}{6} + 2\pi, \quad x = \frac{5\pi}{2}, \quad x = \frac{7\pi}{2}
\]
Solution
Solve the equation by following steps:
- step0: Solve for \(x\):
\(\cos\left(\frac{3\pi }{2}+2x\right)=\cos\left(x\right)\)
- step1: Rewrite the expression:
\(\sin\left(2x\right)=\cos\left(x\right)\)
- step2: Rewrite the expression:
\(2\sin\left(x\right)\cos\left(x\right)=\cos\left(x\right)\)
- step3: Move the expression to the left side:
\(2\sin\left(x\right)\cos\left(x\right)-\cos\left(x\right)=0\)
- step4: Factor the expression:
\(\cos\left(x\right)\left(2\sin\left(x\right)-1\right)=0\)
- step5: Separate into possible cases:
\(\begin{align}&\cos\left(x\right)=0\\&2\sin\left(x\right)-1=0\end{align}\)
- step6: Solve the equation:
\(\begin{align}&x=\frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\\&x=\left\{ \begin{array}{l}\frac{\pi }{6}+2k\pi \\\frac{5\pi }{6}+2k\pi \end{array}\right.,k \in \mathbb{Z}\end{align}\)
- step7: Find the union:
\(x=\left\{ \begin{array}{l}\frac{\pi }{6}+2k\pi \\\frac{\pi }{2}+k\pi \\\frac{5\pi }{6}+2k\pi \end{array}\right.,k \in \mathbb{Z}\)
Solve the equation \( x=\frac{\pi }{6}+2k\pi \).
Solve the equation by following steps:
- step0: Solve for \(k\):
\(x=\frac{\pi }{6}+2k\pi \)
- step1: Multiply the numbers:
\(x=\frac{\pi }{6}+2\pi k\)
- step2: Swap the sides:
\(\frac{\pi }{6}+2\pi k=x\)
- step3: Move the constant to the right side:
\(2\pi k=x-\frac{\pi }{6}\)
- step4: Subtract the terms:
\(2\pi k=\frac{6x-\pi }{6}\)
- step5: Multiply by the reciprocal:
\(2\pi k\times \frac{1}{2\pi }=\frac{6x-\pi }{6}\times \frac{1}{2\pi }\)
- step6: Multiply:
\(k=\frac{6x-\pi }{12\pi }\)
Solve the equation \( x=\frac{5\pi }{6}+2k\pi \).
Solve the equation by following steps:
- step0: Solve for \(k\):
\(x=\frac{5\pi }{6}+2k\pi \)
- step1: Multiply the numbers:
\(x=\frac{5\pi }{6}+2\pi k\)
- step2: Swap the sides:
\(\frac{5\pi }{6}+2\pi k=x\)
- step3: Move the constant to the right side:
\(2\pi k=x-\frac{5\pi }{6}\)
- step4: Subtract the terms:
\(2\pi k=\frac{6x-5\pi }{6}\)
- step5: Multiply by the reciprocal:
\(2\pi k\times \frac{1}{2\pi }=\frac{6x-5\pi }{6}\times \frac{1}{2\pi }\)
- step6: Multiply:
\(k=\frac{6x-5\pi }{12\pi }\)
Solve the equation \( x=\frac{\pi }{2}+k\pi \).
Solve the equation by following steps:
- step0: Solve for \(k\):
\(x=\frac{\pi }{2}+k\pi \)
- step1: Reorder the terms:
\(x=\frac{\pi }{2}+\pi k\)
- step2: Swap the sides:
\(\frac{\pi }{2}+\pi k=x\)
- step3: Move the constant to the right side:
\(\pi k=x-\frac{\pi }{2}\)
- step4: Subtract the terms:
\(\pi k=\frac{2x-\pi }{2}\)
- step5: Multiply by the reciprocal:
\(\pi k\times \frac{1}{\pi }=\frac{2x-\pi }{2}\times \frac{1}{\pi }\)
- step6: Multiply:
\(k=\frac{2x-\pi }{2\pi }\)
Решим уравнение:
### a) Решение уравнения
Дано уравнение:
\[
\cos \left(\frac{3 \pi}{2}+2 x\right)=\cos x
\]
Согласно свойству косинуса, если \(\cos A = \cos B\), то:
\[
A = 2k\pi \pm B, \quad k \in \mathbb{Z}
\]
Подставим \(A = \frac{3\pi}{2} + 2x\) и \(B = x\):
1. \(\frac{3\pi}{2} + 2x = 2k\pi + x\)
2. \(\frac{3\pi}{2} + 2x = 2k\pi - x\)
Решим каждое из этих уравнений.
#### Уравнение 1:
\[
\frac{3\pi}{2} + 2x = 2k\pi + x
\]
Переносим \(x\) влево:
\[
\frac{3\pi}{2} + x = 2k\pi
\]
\[
x = 2k\pi - \frac{3\pi}{2}
\]
#### Уравнение 2:
\[
\frac{3\pi}{2} + 2x = 2k\pi - x
\]
Переносим \(x\) влево:
\[
\frac{3\pi}{2} + 3x = 2k\pi
\]
\[
3x = 2k\pi - \frac{3\pi}{2}
\]
\[
x = \frac{2k\pi - \frac{3\pi}{2}}{3}
\]
Таким образом, корни уравнения:
\[
x = \left\{ \frac{\pi}{6} + 2k\pi, \frac{\pi}{2} + k\pi, \frac{5\pi}{6} + 2k\pi \right\}, \quad k \in \mathbb{Z}
\]
### б) Корни на отрезке \(\left[\frac{5 \pi}{2}; 4 \pi\right]\)
Теперь найдем корни, принадлежащие отрезку \(\left[\frac{5 \pi}{2}; 4 \pi\right]\).
1. **Для \(x = \frac{\pi}{6} + 2k\pi\)**:
\[
\frac{\pi}{6} + 2k\pi \in \left[\frac{5\pi}{2}; 4\pi\right]
\]
Решим для \(k\).
2. **Для \(x = \frac{\pi}{2} + k\pi\)**:
\[
\frac{\pi}{2} + k\pi \in \left[\frac{5\pi}{2}; 4\pi\right]
\]
Решим для \(k\).
3. **Для \(x = \frac{5\pi}{6} + 2k\pi\)**:
\[
\frac{5\pi}{6} + 2k\pi \in \left[\frac{5\pi}{2}; 4\pi\right]
\]
Решим для \(k\).
Теперь найдем значения \(k\) для каждого случая.
### 1. Для \(x = \frac{\pi}{6} + 2k\pi\):
\[
\frac{5\pi}{2} \leq \frac{\pi}{6} + 2k\pi \leq 4\pi
\]
### 2. Для \(x = \frac{\pi}{2} + k\pi\):
\[
\frac{5\pi}{2} \leq \frac{\pi}{2} + k\pi \leq 4\pi
\]
### 3. Для \(x = \frac{5\pi}{6} + 2k\pi\):
\[
\frac{5\pi}{2} \leq \frac{5\pi}{6} + 2k\pi \leq 4\pi
\]
Теперь решим каждое из этих неравенств для \(k\).
Поскольку возникли ошибки при попытке решить неравенства, я выполню их вручную.
### Решение неравенств
1. **Для \(x = \frac{\pi}{6} + 2k\pi\)**:
\[
2k\pi \geq \frac{5\pi}{2} - \frac{\pi}{6} \implies 2k\pi \geq \frac{15\pi}{6} - \frac{\pi}{6} = \frac{14\pi}{6} \implies k \geq \frac{7}{6}
\]
\[
2k\pi \leq 4\pi - \frac{\pi}{6} \implies 2k\pi \leq \frac{24\pi}{6} - \frac{\pi}{6} = \frac{23\pi}{6} \implies k \leq \frac{23}{12}
\]
Значит, \(k = 1\).
2. **Для \(x = \frac{\pi}{2} + k\pi\)**:
\[
k\pi \geq \frac{5\pi}{2} - \frac{\pi}{2} \implies k\pi \geq 2\pi \implies k \geq 2
\]
\[
k\pi \leq 4\pi - \frac{\pi}{2} \implies k\pi \leq \frac{8\pi}{2} - \frac{\pi}{2} = \frac{7\pi}{2} \implies k \leq 3
\]
Значит, \(k = 2\) или \(k = 3\).
3. **Для \(x = \frac{5\pi}{6} + 2k\pi\)**:
\[
2k\pi \geq \frac{5\pi}{2} - \frac{5\pi}{6} \implies 2k\pi \geq \frac{15\pi}{6} - \frac{5\pi}{6} = \frac{10\pi}{6} \implies k \geq \frac{5}{6}
\]
\[
2k\pi \leq 4\pi - \frac{5\pi}{6} \implies 2k\pi \leq \frac{24\pi}{6}
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