13 a) Решите уравнение \[ \cos \left(\frac{3 \pi}{2}+2 x\right)=\cos x \] б) Укажите корни этого уравнения, принадтежащие отрезку \( \left[\frac{5 \pi}{2} ; 4 \pi\right] \).

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}