4cos2x - cosx-3 =0
0 to 2pi
recall cos 2x = 2cos^2 x - 1
4cos2x - cosx-3 =0
4(2cos^2 x - 1 ) - cosx - 3 = 0
8cos^2x - 4 - cosx - 3 = 0
8cos^2 x - cosx - 7 = 0
(cosx - 1)(8cosx + 7) = 0
cosx = 1 or cosx = -7/8
cosx = 1 ---> x = 0 or x = 2π
cosx = -7/8 ---> x = π - cos^-1 (7/8) or x = π + cos^-1 (7/8)
x = appr 2.64 or x = 3.64
let's check one of them
x = 3.64
LS = 4cos(7.28) - cos(3.64) - 3
= .027.. , looks good, close to zero
To solve the equation 4cos(2x) - cos(x) - 3 = 0 between 0 and 2π, one approach is to use algebraic manipulation and trigonometric identities.
Step 1: Notice that we have two different trigonometric functions, cos(2x) and cos(x). We can replace cos(2x) using the double-angle identity for cosine: cos(2x) = 2cos^2(x) - 1.
Step 2: Substituting the identity into the equation, we get:
4(2cos^2(x) - 1) - cos(x) - 3 = 0.
Step 3: Simplify the equation:
8cos^2(x) - 4 - cos(x) - 3 = 0.
8cos^2(x) - cos(x) - 7 = 0.
Step 4: Now we have a quadratic equation in terms of cos(x). We can solve this by factoring, completing the square, or using the quadratic formula.
In this case, factoring is not straightforward, so we can use the quadratic formula to solve for cos(x):
cos(x) = (-b ± √(b² - 4ac))/(2a).
Using a = 8, b = -1, and c = -7, we can substitute these values into the quadratic formula and solve for cos(x).
cos(x) = (-(-1) ± √((-1)² - 4(8)(-7)))/(2(8))
cos(x) = (1 ± √(1 + 224))/16
cos(x) = (1 ± √225)/16.
Step 5: Now we have two possible values for cos(x):
cos(x) = (1 + √225)/16 = (1 + 15)/16 = 16/16 = 1,
or
cos(x) = (1 - √225)/16 = (1 - 15)/16 = -14/16.
Step 6: Recall that cos(x) is equal to the x-coordinate on the unit circle. So, we need to find the corresponding angles in the range from 0 to 2π where the cosine is equal to either 1 or -14/16.
For cos(x) = 1, the corresponding angle is x = 0 radians (0°).
For cos(x) = -14/16, we can use the inverse cosine function to find the angle:
x = arccos(-14/16).
Step 7: Finally, we can find the solutions for x in the range from 0 to 2π (0 to 360°) by substituting the values we found.
The solution x = 0 radians (0°) satisfies the equation.
To find the solution for x = arccos(-14/16) radians, we can use a calculator to find the approximate value of arccos(-14/16), which is approximately 2.943 radians (168.714°).
Therefore, the solutions for the given equation in the range from 0 to 2π (0 to 360°) are x = 0 radians (0°) and x ≈ 2.943 radians (168.714°).