Find all exact solutions on the interval [0,2pi).
cos(8x)+cos(4x)=0
let's use the identity
cos (2A) = 2cos^2 (A) - 1
cos(8x)+cos(4x)=0
2cos^2 (4x) - 1 + cos(4x) = 0
let cos(4x) = y
then we have
2y^2 + y - 1 = 0
(2y - 1)(y + 1) = 0
y = 1/2 or y = -1
if y = 1/2, cos 4x = 1/2
4x = π/3 or 4x = 2π - π/3 = 5π/3
x = π/12 , or x = 5π/12 , (15° or 75°)
but the period of cos(4x) = 2π/4 = π/2 , (90°)
so adding multiples of π/2 will give us more answers.
x = π/12, 7π/12, 13π/12, 19π/12,
in degrees: 15°, 105°, 195°, 285°, 375° <--- the last one is too big
if y = -1
cos 4x = -1
4x = π
x = π/4 , with the same period of cos(4x) to be π/2
x = π/4, 3π/4, 5π/4, 7π/4
in degrees: 45°, 135°, 225°, 315°
To find all exact solutions on the interval [0,2π) for the equation cos(8x) + cos(4x) = 0, we can use trigonometric identities.
First, let's simplify the equation using the double-angle identity for cosine: cos(2A) = 2cos^2(A) - 1.
cos(8x) + cos(4x) = 0
2cos^2(4x) - 1 + cos(4x) = 0
Now, let's substitute a variable to simplify the equation:
Let's replace cos(4x) with a variable, say, u.
2u^2 - 1 + u = 0
2u^2 + u - 1 = 0
Now, we have a quadratic equation in terms of u.
To solve this quadratic equation, we can either factor it or use the quadratic formula. However, factoring is not possible, so we will use the quadratic formula:
u = (-b ± sqrt(b^2 - 4ac)) / 2a
Applying the quadratic formula:
u = (-(1) ± sqrt((1)^2 - 4(2)(-1))) / (2(2))
u = (-1 ± sqrt(1 + 8)) / 4
u = (-1 ± sqrt(9)) / 4
Taking the positive and negative square roots:
u1 = (-1 + 3) / 4 = 1/2
u2 = (-1 - 3) / 4 = -1
Now, let's substitute u back into cos(4x):
For u1:
cos(4x) = 1/2
Using the inverse cosine function:
4x = cos^(-1)(1/2)
There are two values for cos^(-1)(1/2): π/3 and 5π/3.
So, for u1:
4x = π/3 and 4x = 5π/3
Dividing both sides by 4:
x = π/12 and x = 5π/12
For u2:
cos(4x) = -1
Using the inverse cosine function:
4x = cos^(-1)(-1)
The value of cos^(-1)(-1) is π.
So, for u2:
4x = π
Dividing both sides by 4:
x = π/4
Therefore, the exact solutions on the interval [0,2π) for the equation cos(8x) + cos(4x) = 0 are:
x = π/12, 5π/12, and π/4.
To find all exact solutions for the equation cos(8x) + cos(4x) = 0 on the interval [0, 2π), we need to solve for x.
Step 1: Identify the possible values for x
Since we are looking for solutions on the interval [0, 2π), we know that x must be between 0 and 2π.
Step 2: Simplify the equation
Using the trigonometric identity cos(A) + cos(B) = 2cos((A + B)/2)cos((A - B)/2), we can rewrite the equation as:
2cos((8x + 4x)/2)cos((8x - 4x)/2) = 0
2cos(6x)cos(2x) = 0
Step 3: Set each factor equal to zero
To find solutions, we need to consider the two factors separately and set each one equal to zero.
a) cos(6x) = 0
To find the values of x when cos(6x) = 0, we need to find the values of x for which the cosine function is equal to zero. This occurs at angles that are multiples of π/2, so we can set 6x equal to π/2, 3π/2, 5π/2, etc.
6x = π/2, 3π/2, 5π/2, ...
x = π/12, π/2, 5π/12, ...
b) cos(2x) = 0
Similarly, solving for cos(2x) = 0, we find that x = π/4, 3π/4, 5π/4, 7π/4.
Step 4: Combine the solutions
Combining the solutions from both factors, we get:
x = π/12, π/2, 5π/12, π/4, 3π/4, 5π/4, 7π/4
These are all the exact solutions for the equation cos(8x) + cos(4x) = 0 on the interval [0, 2π).