4 cos 2x + 3 cos x = 1
recall that cos 2x = 2cos^2 x - 1
4 cos 2x + 3 cos x = 1
4(2cos^2 x - 1) + 3cosx - 1 = 0
8cos^2 x - 4 + 3cosx - 1 = 0
8cos^2 x + 3cosx - 5 = 0
(cosx + 1)(8cosx - 5) = 0
cosx = -1 or cosx = 5/8
x = 180° or x = 51.32° or x = 308.68°
or, in radians
x = π, or x = .8957 or x = 5.388
To solve the equation 4cos(2x) + 3cos(x) = 1, we can start by simplifying the equation using trigonometric identities.
1. Recall the double angle identity for cosine: cos(2x) = 2cos^2(x) - 1. We can substitute this back into our equation:
4(2cos^2(x) - 1) + 3cos(x) = 1
2. Distribute the 4 to each term inside the parentheses:
8cos^2(x) - 4 + 3cos(x) = 1
3. Rearrange the terms and combine like terms:
8cos^2(x) + 3cos(x) - 5 = 0
Now, we have a quadratic equation in terms of cosine. Let's solve it using the quadratic formula.
4. The quadratic formula is given by: x = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = 8, b = 3, and c = -5. Plugging these values into the formula:
x = (-3 ± √(3^2 - 4(8)(-5))) / (2(8))
5. Simplify the expression inside the square root:
x = (-3 ± √(9 + 160)) / 16
x = (-3 ± √169) / 16
x = (-3 ± 13) / 16
6. Split the equation into two parts for the plus and minus signs:
a) x = (-3 + 13) / 16 = 10 / 16 = 5 / 8
b) x = (-3 - 13) / 16 = -16 / 16 = -1
So the values of x that solve the equation 4cos(2x) + 3cos(x) = 1 are x = 5/8 and x = -1.