Question

6 cos^2x + cos x- 1= 0

Answer 1

6 cos²x + cosx - 1 = 0
cosx = t
6t² + t - 1 = 0
D = 1² + 4 * 6 * 1 = 1 + 24 = 25
t1 = (-1 + 5)/2 * 6 = 4/12 = 1/3
t2 = (-1 - 5)/2 * 6 = -6/12 = -1/2
cosx = -1/2;
x = ±2/3π + 2πk;
cosx = 1/3;
x = arccos1/3 + 2πk;

Answer 2

The given equation is a quadratic equation in terms of the trigonometric function cosine (cos x). To solve for the values of x that satisfy the equation, let's work through the solution step by step:

The equation is: 6cos^2x + cosx - 1 = 0

Let's substitute cos x with a variable, say "u", to make the equation look like a standard quadratic equation:

6u^2 + u - 1 = 0

Now we can solve this quadratic equation for "u" using the quadratic formula:

u = (-b ± √(b^2 - 4ac)) / 2a

In this equation, a = 6, b = 1, and c = -1. Plugging in these values:

u = (-(1) ± √(1^2 - 4 * 6 * (-1))) / (2 * 6) u = (-1 ± √(1 + 24)) / 12 u = (-1 ± √25) / 12 u = (-1 ± 5) / 12

This gives us two possible values for "u":

1. u = (4) / 12 = 1/3
2. u = (-6) / 12 = -1/2

Now, we'll substitute these values of "u" back in terms of cos x:

1. cos x = 1/3
2. cos x = -1/2

To find the corresponding angles, you can use the inverse cosine function (arccos):

1. For cos x = 1/3: x = arccos(1/3) + 2πn or x = -arccos(1/3) + 2πn, where n is an integer.

2. For cos x = -1/2: x = arccos(-1/2) + 2πn or x = -arccos(-1/2) + 2πn, where n is an integer.

These equations will give you the solutions for "x" that satisfy the original equation. Remember that trigonometric equations have periodic solutions due to the periodic nature of trigonometric functions. Therefore, adding multiples of 2π (full circle) to the angle solutions will give you all possible solutions for the given equation.