Let y = cos(x). The equation becomes 4y^3 + 2√3y^2 - 2y - √3 = 0
This is a cubic equation that can be solved using numerical methods or factoring. By inspection, we find that y = cos(π/6) = √3/2 is a root of the equation, making y - √3/2 a factor. We can then use polynomial long division to find the remaining quadratic factor, or use synthetic division to divide the equation by (y - √3/2).
After finding the factors, we can solve for y and then the corresponding x values by taking the arccosine of the roots.
To solve the equation cosx + cos3x + √3cos2x = 0, we can use the trigonometric identity cos(3x) = 4cos^3(x) - 3cos(x) and cos(2x) = 2cos^2(x) - 1.
Substitute these identities into the equation
cos(x) + (4cos^3(x) - 3cos(x)) + √3(2cos^2(x) - 1) =
cos(x) + 4cos^3(x) - 3cos(x) + 2√3cos^2(x) - √3 = 0
Rearranging the terms
4cos^3(x) + 2√3cos^2(x) - 2cos(x) - √3 = 0
Let y = cos(x). The equation becomes
4y^3 + 2√3y^2 - 2y - √3 = 0
This is a cubic equation that can be solved using numerical methods or factoring. By inspection, we find that y = cos(π/6) = √3/2 is a root of the equation, making y - √3/2 a factor. We can then use polynomial long division to find the remaining quadratic factor, or use synthetic division to divide the equation by (y - √3/2).
After finding the factors, we can solve for y and then the corresponding x values by taking the arccosine of the roots.