Find all solutions in the interval [0, 360) 8cos^2x+6cosx+1=0
(4cosx+1)(2cosx+1) = 0
now it's a cinch...
To find all solutions in the given interval, we'll start by solving the quadratic equation 8cos^2x + 6cosx + 1 = 0. Let's solve it step by step.
Step 1: Let's substitute cos(x) with t, so the equation becomes 8t^2 + 6t + 1 = 0.
Step 2: Now we need to factorize the quadratic equation. However, if it cannot be factorized, we'll use the quadratic formula. In this case, we'll go ahead with the quadratic formula.
The quadratic formula is given by x = (-b ± √(b^2 - 4ac)) / (2a).
For our equation, the values are a = 8, b = 6, and c = 1. Plugging these values into the quadratic formula, we get:
t = (-6 ± √(6^2 - 4*8*1)) / (2*8)
t = (-6 ± √(36 - 32)) / 16
t = (-6 ± √4) / 16
t = (-6 ± 2) / 16
Step 3: Simplifying further, we get two possible values for t:
t1 = (-6 + 2) / 16 = -4/16 = -1/4
t2 = (-6 - 2) / 16 = -8/16 = -1/2
Step 4: Since cos(x) = t, we need to find the values of x for which cos(x) equals -1/4 and -1/2.
To find the solutions for cos(x) = -1/4, we can use the inverse cosine function. So we have:
x1 = arccos(-1/4)
And for cos(x) = -1/2, the solution is:
x2 = arccos(-1/2)
Step 5: Now, let's determine the values of x in the interval [0, 360) for which cos(x) equals -1/4 and -1/2. We can use a calculator or a reference table.
Using a calculator, we find:
x1 ≈ 104.48° and 255.52°
x2 ≈ 120° and 240°
Therefore, the solutions in the interval [0, 360) for the given equation are approximately:
x = {104.48°, 120°, 240°, 255.52°}
Note: The values of x can also be expressed in radians, depending on the instructions provided.
To find all solutions for the given equation 8cos^2x + 6cosx + 1 = 0 in the interval [0, 360), we'll follow these steps:
Step 1: Rewrite the equation using a substitution:
Let's substitute cosx with a new variable, say u. So the equation becomes:
8u^2 + 6u + 1 = 0
Step 2: Solve the quadratic equation:
To solve the quadratic equation, we'll use factoring or the quadratic formula. However, in this case, the equation does not factor easily. So we'll use the quadratic formula:
u = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = 8, b = 6, and c = 1. Substituting these values into the quadratic formula gives us:
u = (-6 ± √(6^2 - 4 * 8 * 1)) / (2 * 8)
u = (-6 ± √(36 - 32)) / 16
u = (-6 ± √4) / 16
u = (-6 ± 2) / 16
So we have two possible values for u: (-6 + 2) / 16 and (-6 - 2) / 16.
Simplifying these expressions, we get:
u1 = -4 / 16 = -1/4
u2 = -8 / 16 = -1/2
Step 3: Substitute back cosx for u:
Now we can substitute the values of u back into cosx to obtain the solutions for x:
cosx = -1/4 or cosx = -1/2
Step 4: Find the angles in the given interval:
To find the angles in the interval [0, 360), we can use the inverse cosine function or the unit circle. Let's use the unit circle:
When cosx = -1/4:
On the unit circle, look for the angles where the x-coordinate is -1/4. The angles for the first quadrant are:
x = arccos(-1/4) ≈ 104.47°
Note: There is an angle in the second quadrant with the same x-coordinate, but we only want solutions in the interval [0, 360), so we'll exclude that.
When cosx = -1/2:
On the unit circle, look for the angles where the x-coordinate is -1/2. The angles for the second and third quadrants are:
x = arccos(-1/2) ≈ 120° and x = 360° - arccos(-1/2) ≈ 240°
Therefore, the solutions in the interval [0, 360) are:
x ≈ 104.47°, 120°, 240°
So, the solutions to the equation 8cos^2x + 6cosx + 1 = 0 in the interval [0, 360) are approximately 104.47°, 120°, and 240°.