Question

 2 cos (х - π/4) + √3 = 0.

Answer 1

х=$\sqrt{3$/2cos + $\pi$/4

/ - дробная черта

Answer 2

To solve the equation $2\cos(x - \frac{\pi}{4}) + \sqrt{3} = 0$, follow these steps:

Step 1: Subtract $\sqrt{3}$ from both sides of the equation to isolate the cosine term: $2\cos(x - \frac{\pi}{4}) = -\sqrt{3}$

Step 2: Divide both sides by 2 to isolate the cosine term: $\cos(x - \frac{\pi}{4}) = -\frac{\sqrt{3}}{2}$

Step 3: To find the values of $x$ that satisfy this equation, we need to find the angles whose cosine is $-\frac{\sqrt{3}}{2}$. These angles are usually associated with the angles $\frac{5\pi}{6}$ and $\frac{7\pi}{6}$, which are in the second and third quadrants, respectively.

Step 4: We can write the general solutions for $x$: $x - \frac{\pi}{4} = \frac{5\pi}{6} + 2n\pi \quad \text{or} \quad x - \frac{\pi}{4} = \frac{7\pi}{6} + 2n\pi$

Where $n$ is an integer representing the number of full rotations (360 degrees or $2\pi$ radians) made.

Step 5: Solve for $x$ in both equations: $x = \frac{5\pi}{6} + \frac{\pi}{4} + 2n\pi \quad \text{or} \quad x = \frac{7\pi}{6} + \frac{\pi}{4} + 2n\pi$

Step 6: Combine like terms: $x = \frac{9\pi}{12} + 2n\pi \quad \text{or} \quad x = \frac{11\pi}{12} + 2n\pi$

Step 7: Simplify further: $x = \frac{3\pi}{4} + 2n\pi \quad \text{or} \quad x = \frac{11\pi}{12} + 2n\pi$

These are the general solutions for $x$ that satisfy the given equation.

Please note that trigonometric equations can have an infinite number of solutions due to the periodic nature of trigonometric functions. The solutions provided above cover all possible values of $x$ that satisfy the equation.