Question

.sin(2x-π/3)=√3/2 .cos(-3x)=√2/2​

Answer 1

Ответ:

решение на фотографии

Answer 2

To solve the trigonometric equations:

1. $\sin(2x - \frac{\pi}{3}) = \frac{\sqrt{3}}{2}$
2. $\cos(-3x) = \frac{\sqrt{2}}{2}$

Let's solve them one by one:

1. $\sin(2x - \frac{\pi}{3}) = \frac{\sqrt{3}}{2}$:

First, find the values of $2x - \frac{\pi}{3}$ that satisfy this equation. Take the inverse sine (arcsin) of both sides:

$2x - \frac{\pi}{3} = \arcsin\left(\frac{\sqrt{3}}{2}\right)$

Now, solve for $x$:

$2x = \arcsin\left(\frac{\sqrt{3}}{2}\right) + \frac{\pi}{3}$

$x = \frac{1}{2}\left(\arcsin\left(\frac{\sqrt{3}}{2}\right) + \frac{\pi}{3}\right)$

Now, calculate the value of $x$:

$x \approx \frac{1}{2}\left(\frac{\pi}{3} + \frac{\pi}{3}\right)$

$x \approx \frac{\pi}{3}$

So, one solution to the equation is $x \approx \frac{\pi}{3}$.

2. $\cos(-3x) = \frac{\sqrt{2}}{2}$:

First, find the values of $-3x$ that satisfy this equation. Take the inverse cosine (arccos) of both sides:

$-3x = \arccos\left(\frac{\sqrt{2}}{2}\right)$

Now, solve for $x$:

$x = -\frac{1}{3}\arccos\left(\frac{\sqrt{2}}{2}\right)$

Now, calculate the value of $x$:

$x \approx -\frac{1}{3}\left(\frac{\pi}{4}\right)$

$x \approx -\frac{\pi}{12}$

So, one solution to the second equation is $x \approx -\frac{\pi}{12}$.

Therefore, the two solutions to the system of equations are $x \approx \frac{\pi}{3}$ and $x \approx -\frac{\pi}{12}$.