Question

Cos(2x+pi/3)*sin(4x-pi/4)=0

Answer 1

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Answer 2

To solve the trigonometric equation $\cos(2x+\frac{\pi}{3}) \cdot \sin(4x-\frac{\pi}{4}) = 0$, we can find the values of $x$ that make either the cosine or the sine factor equal to zero.

For the cosine factor to be zero, we have $\cos(2x+\frac{\pi}{3}) = 0$. Solving for $x$:

$2x + \frac{\pi}{3} = \frac{\pi}{2} + k\pi, \quad \text{where } k \text{ is an integer}$

$2x = \frac{\pi}{2} - \frac{\pi}{3} + k\pi$

$x = \frac{\pi}{6} + \frac{k\pi}{2}$

For the sine factor to be zero, we have $\sin(4x-\frac{\pi}{4}) = 0$. Solving for $x$:

$4x - \frac{\pi}{4} = k\pi, \quad \text{where } k \text{ is an integer}$

$4x = \frac{\pi}{4} + k\pi$

$x = \frac{\pi}{16} + \frac{k\pi}{4}$

So, the general solutions for $x$ that satisfy the equation are:

$x = \frac{\pi}{6} + \frac{k\pi}{2}, \quad \text{or} \quad x = \frac{\pi}{16} + \frac{k\pi}{4}$

Where $k$ is any integer. These solutions cover all the values of $x$ for which the given trigonometric equation holds true.