I) $$ \cos 8 x+\cos 4 x=2 \cos 2 x-4 \operatorname{sen}^{2} 3 x \cdot \cos 2 x $$

Question

UpStudy AI · Accepted Answer

Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\cos\left(8x\right)+\cos\left(4x\right)=2\cos\left(2x\right)-4\sin^{2}\left(3x\right)\cos\left(2x\right)$$
- step1: Transform the expression:
  $$2\cos\left(6x\right)\cos\left(2x\right)=2\cos\left(2x\right)-4\sin^{2}\left(3x\right)\cos\left(2x\right)$$
- step2: Rewrite the expression:
  $$2\cos\left(2x\right)\cos\left(6x\right)=-2\cos\left(2x\right)+4\cos\left(2x\right)\cos^{2}\left(3x\right)$$
- step3: Move the expression to the left side:
  $$2\cos\left(2x\right)\cos\left(6x\right)-\left(-2\cos\left(2x\right)+4\cos\left(2x\right)\cos^{2}\left(3x\right)\right)=0$$
- step4: Calculate:
  $$2\cos\left(2x\right)\cos\left(6x\right)+2\cos\left(2x\right)-4\cos\left(2x\right)\cos^{2}\left(3x\right)=0$$
- step5: Factor the expression:
  $$2\cos\left(2x\right)\left(\cos\left(6x\right)+1-2\cos^{2}\left(3x\right)\right)=0$$
- step6: Calculate:
  $$\cos\left(2x\right)\left(\cos\left(6x\right)+1-2\cos^{2}\left(3x\right)\right)=0$$
- step7: Separate into possible cases:
  $$\begin{align}&\cos\left(2x\right)=0\&\cos\left(6x\right)+1-2\cos^{2}\left(3x\right)=0\end{align}$$
- step8: Solve the equation:
  $$\begin{align}&x=\frac{\pi }{4}+\frac{k\pi }{2},k \in \mathbb{Z}\&x \in \mathbb{R}\end{align}$$
- step9: Find the union:
  $$x \in \mathbb{R}$$
The equation

$$
\cos 8x + \cos 4x = 2 \cos 2x - 4 \sin^2 3x \cdot \cos 2x
$$

has the solution:

$$
x \in \mathbb{R}
$$

This means that all real numbers are solutions to the equation.
All real numbers are solutions to the equation.

I)

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