a) Putting 5 cos²x+4 cosx = 1 in standard quadratic trigonometric equation form:To put the equation in standard quadratic trigonometric equation form, move all the terms to the left-hand side of the equation and simplify it. Therefore, 5 cos²x + 4 cosx - 1 = 0 is the standard quadratic trigonometric equation form.b) Using the quadratic formula to factor the equation.The quadratic formula is utilized to factor the equation. For any quadratic equation of the form ax²+ bx + c = 0, the quadratic formula is x = {-b ± √(b² - 4ac)} / 2a.5 cos²x + 4 cosx - 1 = 0 is the equation that we need to factor.Now substitute a = 5, b = 4, and c = -1 in the quadratic formula.x = {-4 ± √(4² - 4(5)(-1))} / 2(5)x = {-4 ± √(16 + 20)} / 10x = {-4 ± √36} / 10x = {-4 ± 6} / 10Therefore, x = -1 or x = 1/5 are the roots of the equation. c) Solutions to the equation to two decimal places, where 0° ≤ x ≤ 360°To find the value of x between 0 and 360 degrees, we must use an inverse trigonometric function.Cosine inverse is the inverse of cosine function that gives us the angle whose cosine is a given value. Cos -1 or arccos is used to denote the inverse of cosine function. cos -1 (1/5) = 78.46 and cos -1 (-1) = 180°, because the cosine function is negative between 90° and 270°. Therefore, the solution to the equation 5 cos²x + 4 cosx - 1 = 0, where 0° ≤ x ≤ 360°, is 78.46°.