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 $$ \left[-5 \pi ;-\frac{7 \pi}{2}\right] $$.

Question

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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 $$[-5\pi; -\frac{7\pi}{2}]$$:

$$
x = -\frac{9\pi}{2}, -\frac{7\pi}{2}, -\frac{10\pi}{3}, -\frac{14\pi}{3}
$$
Корни уравнения на отрезке $$[-5\pi; -\frac{7\pi}{2}]$$ равны: $$ x = -\frac{9\pi}{2}, -\frac{7\pi}{2}, -\frac{10\pi}{3}, -\frac{14\pi}{3} $$

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 \( \left[-5 \pi ;-\frac{7 \pi}{2}\right] \).

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