Question

2cos2(6x)+8sin(3x)cos(3x)-4=0​

Answer 1

$2cos^2(6x)+8sin(3x)cos(3x)-4=0\\2(1-sin^2(6x))+4sin(6x)-4=0\\sin(6x)=t; -1\leq t\leq 1\\2(1-t^2)+4t-4=0 |:(-2)\\-1+t^2-2t+2=0\\t^2-2t+1=0\\(t-1)^2=0\\t-1=0\\t=1\\sin(6x)=1\\6x=\frac{\pi}{2} +2\pi k, k \in Z\\x=\frac{\pi}{12} +\frac{\pi k}{3}, k \in Z$

Answer 2

To solve the trigonometric equation $2\cos^2(6x) + 8\sin(3x)\cos(3x) - 4 = 0$, we can use some trigonometric identities to simplify and then solve for $x$.

First, let's rewrite the equation using trigonometric identities:

$$2\cos^2(6x) + 8\sin(3x)\cos(3x) - 4 = 0$$

We can use the double angle formula for $\cos(2\theta)$:

$$\cos(2\theta) = 2\cos^2(\theta) - 1$$

Now, let's substitute $\cos^2(6x)$ with $\frac{1}{2}\cos(12x) + \frac{1}{2}$:

$$2\left(\frac{1}{2}\cos(12x) + \frac{1}{2}\right) + 8\sin(3x)\cos(3x) - 4 = 0$$

Simplify:

$$\cos(12x) + 4\sin(3x)\cos(3x) - 4 = 0$$

Now, let's use the double angle formula for $\sin(2\theta)$:

$$\sin(2\theta) = 2\sin(\theta)\cos(\theta)$$

Substitute $\sin(3x)\cos(3x)$ with $\frac{1}{2}\sin(6x)$:

$$\cos(12x) + 4\left(\frac{1}{2}\sin(6x)\right) - 4 = 0$$

Simplify further:

$$\cos(12x) + 2\sin(6x) - 4 = 0$$

Now, you have a trigonometric equation in terms of $\cos$ and $\sin$. You can use various trigonometric identities and algebraic techniques to solve for $x$. Unfortunately, this equation doesn't have a simple closed-form solution, and you may need to use numerical methods or calculators to find approximate solutions for $x$.