To solve the equation 2cos3x−π/63x - π/63x−π/6 + √3 = 0, we can first isolate the cosine term:
2cos3x−π/63x - π/63x−π/6 = -√3
Next, divide both sides by 2:
cos3x−π/63x - π/63x−π/6 = -√3 / 2
Now, we want to find the values of x that satisfy this equation. Since we have a cosine function and we know that the cosine function is periodic with a period of 2π, we can say that:
3x - π/6 = ±π/6 + 2πn
Where n is an integer. Solving for x, we get:
3x = ±π/6 + π/6 + 2πn
3x = π/3 + 2πn
x = π/9 + 2πn/3
So, the solution to the equation 2cos3x−π/63x - π/63x−π/6 + √3 = 0 is:
To solve the equation 2cos3x−π/63x - π/63x−π/6 + √3 = 0, we can first isolate the cosine term:
2cos3x−π/63x - π/63x−π/6 = -√3
Next, divide both sides by 2:
cos3x−π/63x - π/63x−π/6 = -√3 / 2
Now, we want to find the values of x that satisfy this equation. Since we have a cosine function and we know that the cosine function is periodic with a period of 2π, we can say that:
3x - π/6 = ±π/6 + 2πn
Where n is an integer. Solving for x, we get:
3x = ±π/6 + π/6 + 2πn
3x = π/3 + 2πn
x = π/9 + 2πn/3
So, the solution to the equation 2cos3x−π/63x - π/63x−π/6 + √3 = 0 is:
x = π/9 + 2πn/3, where n is an integer.