a) Решите уравнение \( \cos x+2 \cos \left(2 x-\frac{\pi}{3}\right)=\sqrt{3} \sin 2 x-1 \) б) Найдите все корни этого уравнения, принадлежашие отрезку \( \left[-5 \pi ;-\frac{7 \pi}{2}\right] \).

Solve the equation by following steps: - step0: Solve for \(x\): \(\cos\left(x\right)+2\cos\left(2x-\frac{\pi }{3}\right)=\sqrt{3}\times \sin\left(2x\right)-1\) - step1: Use the periodicity identities: \(\cos\left(x\right)+2\cos\left(2x+\frac{5\pi }{3}\right)=\sqrt{3}\times \sin\left(2x\right)-1\) - step2: Move the expression to the left side: \(\cos\left(x\right)+2\cos\left(2x+\frac{5\pi }{3}\right)-\left(\sqrt{3}\times \sin\left(2x\right)-1\right)=0\) - step3: Calculate: \(\cos\left(x\right)+2\cos\left(2x+\frac{5\pi }{3}\right)-\sqrt{3}\times \sin\left(2x\right)+1=0\) - step4: Simplify: \(\cos\left(x\right)+\cos^{2}\left(x\right)-\sin^{2}\left(x\right)+1=0\) - step5: Solve using substitution: \(\frac{1-\tan^{2}\left(\frac{1}{2}x\right)}{1+\tan^{2}\left(\frac{1}{2}x\right)}+\left(\frac{1-\tan^{2}\left(\frac{1}{2}x\right)}{1+\tan^{2}\left(\frac{1}{2}x\right)}\right)^{2}-\left(\frac{2\tan\left(\frac{1}{2}x\right)}{1+\tan^{2}\left(\frac{1}{2}x\right)}\right)^{2}+1=0\) - step6: Solve using substitution: \(\frac{1-t^{2}}{1+t^{2}}+\left(\frac{1-t^{2}}{1+t^{2}}\right)^{2}-\left(\frac{2t}{1+t^{2}}\right)^{2}+1=0\) - step7: Multiply both sides of the equation by LCD: \(\left(\frac{1-t^{2}}{1+t^{2}}+\left(\frac{1-t^{2}}{1+t^{2}}\right)^{2}-\left(\frac{2t}{1+t^{2}}\right)^{2}+1\right)\left(1+2t^{2}+t^{4}\right)=0\times \left(1+2t^{2}+t^{4}\right)\) - step8: Simplify the equation: \(3-4t^{2}+t^{4}=0\) - step9: Factor the expression: \(\left(1-t\right)\left(1+t\right)\left(3-t^{2}\right)=0\) - step10: Separate into possible cases: \(\begin{align}&1-t=0\\&1+t=0\\&3-t^{2}=0\end{align}\) - step11: Solve the equation: \(\begin{align}&t=1\\&t=-1\\&t=\sqrt{3}\\&t=-\sqrt{3}\end{align}\) - step12: Substitute back: \(\begin{align}&\tan\left(\frac{1}{2}x\right)=1\\&\tan\left(\frac{1}{2}x\right)=-1\\&\tan\left(\frac{1}{2}x\right)=\sqrt{3}\\&\tan\left(\frac{1}{2}x\right)=-\sqrt{3}\end{align}\) - step13: Calculate: \(\begin{align}&x=\frac{\pi }{2}+2k\pi ,k \in \mathbb{Z}\\&\tan\left(\frac{1}{2}x\right)=-1\\&\tan\left(\frac{1}{2}x\right)=\sqrt{3}\\&\tan\left(\frac{1}{2}x\right)=-\sqrt{3}\end{align}\) - step14: Calculate: \(\begin{align}&x=\frac{\pi }{2}+2k\pi ,k \in \mathbb{Z}\\&x=\frac{3\pi }{2}+2k\pi ,k \in \mathbb{Z}\\&\tan\left(\frac{1}{2}x\right)=\sqrt{3}\\&\tan\left(\frac{1}{2}x\right)=-\sqrt{3}\end{align}\) - step15: Calculate: \(\begin{align}&x=\frac{\pi }{2}+2k\pi ,k \in \mathbb{Z}\\&x=\frac{3\pi }{2}+2k\pi ,k \in \mathbb{Z}\\&x=\frac{2\pi }{3}+2k\pi ,k \in \mathbb{Z}\\&\tan\left(\frac{1}{2}x\right)=-\sqrt{3}\end{align}\) - step16: Calculate: \(\begin{align}&x=\frac{\pi }{2}+2k\pi ,k \in \mathbb{Z}\\&x=\frac{3\pi }{2}+2k\pi ,k \in \mathbb{Z}\\&x=\frac{2\pi }{3}+2k\pi ,k \in \mathbb{Z}\\&x=\frac{4\pi }{3}+2k\pi ,k \in \mathbb{Z}\end{align}\) - step17: Check if \(x=\pi+2k\pi,k\in\mathbb{Z}\) is a solution\(:\) \(\cos\left(\pi +2k\pi\right)+2\cos\left(2\left(\pi +2k\pi\right)-\frac{\pi }{3}\right)=\sqrt{3}\times \sin\left(2\left(\pi +2k\pi\right)\right)-1\) - step18: Calculate: \(\cos\left(\pi \right)+2\cos\left(2\pi -\frac{\pi }{3}\right)=\sqrt{3}\times \sin\left(2\pi \right)-1\) - step19: Simplify: \(0=-1\) - step20: Check the equality: \(\textrm{false}\) - step21: Since \(x=\pi+2k\pi,k\in\mathbb{Z}\) is not a solution,don't include it\(:\) \(\begin{align}&x=\frac{\pi }{2}+2k\pi ,k \in \mathbb{Z}\\&x=\frac{3\pi }{2}+2k\pi ,k \in \mathbb{Z}\\&x=\frac{2\pi }{3}+2k\pi ,k \in \mathbb{Z}\\&x=\frac{4\pi }{3}+2k\pi ,k \in \mathbb{Z}\end{align}\) - step22: Find the union: \(x=\left\{ \begin{array}{l}\frac{\pi }{2}+k\pi \\\frac{2\pi }{3}+2k\pi \\\frac{4\pi }{3}+2k\pi \end{array}\right.,k \in \mathbb{Z}\) 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: \(k=\frac{2x-\pi }{2\pi }\) Solve the equation \( x=\frac{2\pi }{3}+2k\pi \). Solve the equation by following steps: - step0: Solve for \(k\): \(x=\frac{2\pi }{3}+2k\pi \) - step1: Multiply the numbers: \(x=\frac{2\pi }{3}+2\pi k\) - step2: Swap the sides: \(\frac{2\pi }{3}+2\pi k=x\) - step3: Move the constant to the right side: \(2\pi k=x-\frac{2\pi }{3}\) - step4: Subtract the terms: \(2\pi k=\frac{3x-2\pi }{3}\) - step5: Multiply by the reciprocal: \(2\pi k\times \frac{1}{2\pi }=\frac{3x-2\pi }{3}\times \frac{1}{2\pi }\) - step6: Multiply: \(k=\frac{3x-2\pi }{6\pi }\) Solve the equation \( x=\frac{4\pi }{3}+2k\pi \). Solve the equation by following steps: - step0: Solve for \(k\): \(x=\frac{4\pi }{3}+2k\pi \) - step1: Multiply the numbers: \(x=\frac{4\pi }{3}+2\pi k\) - step2: Swap the sides: \(\frac{4\pi }{3}+2\pi k=x\) - step3: Move the constant to the right side: \(2\pi k=x-\frac{4\pi }{3}\) - step4: Subtract the terms: \(2\pi k=\frac{3x-4\pi }{3}\) - step5: Multiply by the reciprocal: \(2\pi k\times \frac{1}{2\pi }=\frac{3x-4\pi }{3}\times \frac{1}{2\pi }\) - step6: Multiply: \(k=\frac{3x-4\pi }{6\pi }\) Solve the equation \( k\pi + \frac{\pi}{2} = -5\pi \). Solve the equation by following steps: - step0: Solve for \(k\): \(k\pi +\frac{\pi }{2}=-5\pi \) - step1: Reorder the terms: \(\pi k+\frac{\pi }{2}=-5\pi \) - step2: Move the constant to the right side: \(\pi k=-5\pi -\frac{\pi }{2}\) - step3: Subtract the numbers: \(\pi k=-\frac{11\pi }{2}\) - step4: Multiply by the reciprocal: \(\pi k\times \frac{1}{\pi }=-\frac{11\pi }{2}\times \frac{1}{\pi }\) - step5: Multiply: \(k=-\frac{11}{2}\) Solve the equation \( k\pi + \frac{\pi}{2} = -\frac{7\pi}{2} \). Solve the equation by following steps: - step0: Solve for \(k\): \(k\pi +\frac{\pi }{2}=-\frac{7\pi }{2}\) - step1: Reorder the terms: \(\pi k+\frac{\pi }{2}=-\frac{7\pi }{2}\) - step2: Move the constant to the right side: \(\pi k=-\frac{7\pi }{2}-\frac{\pi }{2}\) - step3: Subtract the numbers: \(\pi k=-4\pi \) - step4: Divide both sides: \(\frac{\pi k}{\pi }=\frac{-4\pi }{\pi }\) - step5: Divide the numbers: \(k=-4\) Solve the equation \( 2k\pi + \frac{4\pi}{3} = -5\pi \). Solve the equation by following steps: - step0: Solve for \(k\): \(2k\pi +\frac{4\pi }{3}=-5\pi \) - step1: Multiply the numbers: \(2\pi k+\frac{4\pi }{3}=-5\pi \) - step2: Move the constant to the right side: \(2\pi k=-5\pi -\frac{4\pi }{3}\) - step3: Subtract the numbers: \(2\pi k=-\frac{19\pi }{3}\) - step4: Multiply by the reciprocal: \(2\pi k\times \frac{1}{2\pi }=-\frac{19\pi }{3}\times \frac{1}{2\pi }\) - step5: Multiply: \(k=-\frac{19}{6}\) Solve the equation \( 2k\pi + \frac{2\pi}{3} = -\frac{7\pi}{2} \). Solve the equation by following steps: - step0: Solve for \(k\): \(2k\pi +\frac{2\pi }{3}=-\frac{7\pi }{2}\) - step1: Multiply the numbers: \(2\pi k+\frac{2\pi }{3}=-\frac{7\pi }{2}\) - step2: Move the constant to the right side: \(2\pi k=-\frac{7\pi }{2}-\frac{2\pi }{3}\) - step3: Subtract the numbers: \(2\pi k=-\frac{25\pi }{6}\) - step4: Multiply by the reciprocal: \(2\pi k\times \frac{1}{2\pi }=-\frac{25\pi }{6}\times \frac{1}{2\pi }\) - step5: Multiply: \(k=-\frac{25}{12}\) Solve the equation \( 2k\pi + \frac{4\pi}{3} = -\frac{7\pi}{2} \). Solve the equation by following steps: - step0: Solve for \(k\): \(2k\pi +\frac{4\pi }{3}=-\frac{7\pi }{2}\) - step1: Multiply the numbers: \(2\pi k+\frac{4\pi }{3}=-\frac{7\pi }{2}\) - step2: Move the constant to the right side: \(2\pi k=-\frac{7\pi }{2}-\frac{4\pi }{3}\) - step3: Subtract the numbers: \(2\pi k=-\frac{29\pi }{6}\) - step4: Multiply by the reciprocal: \(2\pi k\times \frac{1}{2\pi }=-\frac{29\pi }{6}\times \frac{1}{2\pi }\) - step5: Multiply: \(k=-\frac{29}{12}\) Solve the equation \( 2k\pi + \frac{2\pi}{3} = -5\pi \). Solve the equation by following steps: - step0: Solve for \(k\): \(2k\pi +\frac{2\pi }{3}=-5\pi \) - step1: Multiply the numbers: \(2\pi k+\frac{2\pi }{3}=-5\pi \) - step2: Move the constant to the right side: \(2\pi k=-5\pi -\frac{2\pi }{3}\) - step3: Subtract the numbers: \(2\pi k=-\frac{17\pi }{3}\) - step4: Multiply by the reciprocal: \(2\pi k\times \frac{1}{2\pi }=-\frac{17\pi }{3}\times \frac{1}{2\pi }\) - step5: Multiply: \(k=-\frac{17}{6}\) ### Решение уравнения a) Мы имеем уравнение: \[ \cos x + 2 \cos \left(2 x - \frac{\pi}{3}\right) = \sqrt{3} \sin 2 x - 1 \] Решение этого уравнения даёт следующие корни: \[ x = \left\{ \begin{array}{l} \frac{\pi }{2} + k\pi \\ \frac{2\pi }{3} + 2k\pi \\ \frac{4\pi }{3} + 2k\pi \end{array} \right., \quad k \in \mathbb{Z} \] ### Поиск корней на заданном отрезке Теперь найдем все корни, принадлежащие отрезку \([-5\pi; -\frac{7\pi}{2}]\). 1. **Для первого корня**: \(x = \frac{\pi}{2} + k\pi\) - Решим уравнения: - \(\frac{\pi}{2} + k\pi = -5\pi\) - \(\frac{\pi}{2} + k\pi = -\frac{7\pi}{2}\) Решения: - \(k = -\frac{11}{2}\) или \(k = -5.5\) - \(k = -4\) 2. **Для второго корня**: \(x = \frac{2\pi}{3} + 2k\pi\) - Решим уравнения: - \(\frac{2\pi}{3} + 2k\pi = -5\pi\) - \(\frac{2\pi}{3} + 2k\pi = -\frac{7\pi}{2}\) Решения: - \(k = -\frac{17}{6}\) или \(k = -2.83\) - \(k = -\frac{25}{12}\) или \(k = -2.08\) 3. **Для третьего корня**: \(x = \frac{4\pi}{3} + 2k\pi\) - Решим уравнения: - \(\frac{4\pi}{3} + 2k\pi = -5\pi\) - \(\frac{4\pi}{3} + 2k\pi = -\frac{7\pi}{2}\) Решения: - \(k = -\frac{19}{6}\) или \(k = -3.16\) - \(k = -\frac{29}{12}\) или \(k = -2.41\) ### Итоговые корни Теперь подставим найденные значения \(k\) обратно в уравнения для \(x\): 1. Для \(k = -5\): \[ x = \frac{\pi}{2} - 5\pi = -\frac{9\pi}{2} \] 2. Для \(k = -4\): \[ x = \frac{\pi}{2} - 4\pi = -\frac{7\pi}{2} \] 3. Для \(k = -2\): \[ x = \frac{2\pi}{3} - 4\pi = -\frac{10\pi}{3} \] 4. Для \(k = -3\): \[ x = \frac{4\pi}{3} - 6\pi = -\frac{14\pi}{3} \] Таким образом, корни уравнения, принадлежащие отрезку \([-5\pi; -\frac{7\pi}{2}]\): \[ x = -\frac{9\pi}{2}, -\frac{7\pi}{2}, -\frac{10\pi}{3}, -\frac{14\pi}{3} \]