Solve the equation in the given domain
cos 4x + cos 2x = 0, 0<(or equal to) x < pi
use the identity
cos 2A = 2 cos^2 A - 1 on cos4A
cos 4A = 2cos^2 2A - 1
so
2cos^2 2A - 1 + cos 2A = 0
(2cos 2A -1)(cos 2A + 1) = 0
cos 2A = 1/2 or cos 2A = -1
2A = 60° or 300° or 2A = 180°
A = 30° or 150° or A = 90°
I guess I should have kept x instead of A, no big deal.
x = 30°, 90° , 150°
or
x = π/6, π/2, 5π/6
check:
if x=30° --- cos 120° + cos 60° = -1/2 + 1/2 = 0
if x=90 --- cos 360 + cos 180 = 1 - 1 =0
if x = 150 -- cos 600 + cos 300 = -1/2 + 1/2 = 0
all is good.
Thanks. Makes sense. My problem was I was trying to use another double argument property.
To solve the equation cos 4x + cos 2x = 0 in the given domain 0 ≤ x < π, we can use the properties of the cosine function and trigonometric identities.
First, let's simplify the equation by applying the double angle formula for cosine: cos(2θ) = 2cos²(θ) - 1.
Using this formula, we can rewrite the equation as:
cos(2x) * (2cos²(2x) - 1) = 0
Now, we have two cases to consider:
Case 1: cos(2x) = 0
If cos(2x) = 0, then 2x = π/2, 3π/2, 5π/2, ...
Simplifying, we have:
x = π/4, 3π/4, 5π/4, ...
However, we need to check if these solutions lie within the given domain 0 ≤ x < π. Only x = π/4 and 3π/4 satisfy this condition.
Case 2: 2cos²(2x) - 1 = 0
If 2cos²(2x) - 1 = 0, then cos²(2x) = 1/2.
Taking the square root of both sides, we get:
cos(2x) = ±√2/2
Using the values from the unit circle, we know that cos(π/4) = √2/2 and cos(7π/4) = √2/2.
Therefore, we have:
2x = π/4, 7π/4
Simplifying, we get:
x = π/8, 7π/8
Again, we need to check if these solutions satisfy the given domain. Only x = π/8 satisfies 0 ≤ x < π.
So, the solutions to the equation cos 4x + cos 2x = 0 in the given domain 0 ≤ x < π are:
x = π/4, 3π/4, π/8
To solve the equation cos(4x) + cos(2x) = 0 in the given domain 0 ≤ x < π, we can use the properties of cosine and algebraic manipulation to find the solutions.
Here's how you can solve the equation step by step:
Step 1: Apply the sum-to-product formula for cosine.
The sum-to-product formula states that cos(A) + cos(B) = 2cos((A + B)/2)cos((A - B)/2).
In our equation, cos(4x) + cos(2x) = 0, we can rewrite it as:
2cos(3x)cos(x) = 0.
Step 2: Set each factor equal to zero.
cos(3x) = 0, and cos(x) = 0.
Step 3: Solve the first equation, cos(3x) = 0.
In the given domain, 0 ≤ x < π, the solutions for cos(3x) = 0 are:
3x = π/2, and 3x = 3π/2.
Dividing both sides by 3 gives:
x = π/6, and x = π/2.
Step 4: Solve the second equation, cos(x) = 0.
In the given domain, 0 ≤ x < π, the solutions for cos(x) = 0 are:
x = π/2.
Step 5: Check if the solutions are within the given domain.
We have found x = π/6 and x = π/2 as potential solutions. Let's check if they satisfy the given domain condition 0 ≤ x < π.
For x = π/6:
0 ≤ π/6 < π (satisfied)
For x = π/2:
0 ≤ π/2 < π (satisfied)
Step 6: Finalize the solutions.
The solution to the equation cos(4x) + cos(2x) = 0 in the given domain 0 ≤ x < π is:
x = π/6, and x = π/2.
So, the solutions are x = π/6 and x = π/2.