Question

& sin (2x-П/4)+1=07​

Answer 1

$\sin \left (2x - \dfrac{\pi}{4} \right)+ 1 = 0$

$\sin \left (2x - \dfrac{\pi}{4} \right) = -1$

$2x - \dfrac{\pi}{4} = -\dfrac{\pi}{2} + 2\pi n, \ n \in \mathbb{Z}$

$2x= \dfrac{\pi}{4} - \dfrac{\pi}{2} + 2\pi n, \ n \in \mathbb{Z}$

$2x = -\dfrac{\pi}{4} + 2\pi n, \ n \in \mathbb{Z} \ \ \ | : 2$

$x = -\dfrac{\pi}{8} + \pi n, \ n \in \mathbb{Z}$

Ответ: $x = -\dfrac{\pi}{8} + \pi n, \ n \in \mathbb{Z}$

Answer 2

To solve the equation &sin(2x-π/4)+1=0.7, we can start by isolating the sine term on one side:

&sin(2x-π/4) = 0.7 - 1 &sin(2x-π/4) = -0.3

Next, we can use the inverse sine function (arcsin or sin^-1) to solve for 2x-π/4:

2x-π/4 = arcsin(-0.3)

Now, we can solve for 2x by adding π/4 to both sides:

2x = arcsin(-0.3) + π/4

Finally, we can solve for x by dividing both sides by 2:

x = (arcsin(-0.3) + π/4) / 2

Please note that the solution will depend on the interval or domain in which you are looking for solutions. Also, keep in mind that the inverse sine function typically returns values in radians.