Question

1)cos(x-П/4)=1 2)sin(x/2+П)=-1/2 3)2sin*3x=✓3 Help

Answer 1

1)cos(x-П/4)=1
2)sin(x/2+П)=-1/2 
3)2sin3x=✓3

Answer 2

Let's solve each trigonometric equation step by step:

1. $\cos(x - \frac{\pi}{4}) = 1$

   To solve this, we need to find the angles $x$ that make $\cos(x - \frac{\pi}{4})$ equal to 1.

   The cosine function is 1 when the angle is 0 degrees (or $2\pi$ radians) or any multiple of $2\pi$. So, we can set $x - \frac{\pi}{4} = 2\pi n$ where $n$ is an integer.

   Solving for $x$:

   $x - \frac{\pi}{4} = 2\pi n$

   $x = 2\pi n + \frac{\pi}{4}$ where $n$ is an integer.

2. $\sin(\frac{x}{2} + \pi) = -\frac{1}{2}$

   To solve this, we need to find the angles $x$ that make $\sin(\frac{x}{2} + \pi)$ equal to $-\frac{1}{2}$.

   The sine function is $-\frac{1}{2}$ at $-\frac{5\pi}{6}$ and $-\frac{7\pi}{6}$, and we want to find $x$ such that $\frac{x}{2} + \pi = -\frac{5\pi}{6} + 2k\pi$ or $\frac{x}{2} + \pi = -\frac{7\pi}{6} + 2k\pi$ where $k$ is an integer.

   Solving for $x$:

   $\frac{x}{2} = -\frac{5\pi}{6} + 2k\pi - \pi$ or $\frac{x}{2} = -\frac{7\pi}{6} + 2k\pi - \pi$

   $x = -\frac{5\pi}{3} + 4k\pi$ or $x = -\frac{11\pi}{3} + 4k\pi$ where $k$ is an integer.

3. $2\sin(3x) = \sqrt{3}$

   To solve this, we need to find the angles $x$ that make $2\sin(3x)$ equal to $\sqrt{3}$.

   Divide both sides by 2:

   $\sin(3x) = \frac{\sqrt{3}}{2}$

   The sine function is $\frac{\sqrt{3}}{2}$ at $\frac{\pi}{3}$ and $\frac{5\pi}{3}$, and we want to find $x$ such that $3x = \frac{\pi}{3} + 2k\pi$ or $3x = \frac{5\pi}{3} + 2k\pi$ where $k$ is an integer.

   Solving for $x$:

   $x=\frac{\pi }{9}+\frac{2}{}$