Find all solutions to the equation in the interval [0, 2pi)
cos4x-cos2x=0
So,this is what i've done so far:
cos4x-cos2x=0
cos2(2x)-cos2x
(2cos^2(2x)-1)-(2cos^2(x)-1)
No idea what to do next.
http://answers.yahoo.com/question/index?qid=20080413170208AAL5ESb
To go further, you can use the trigonometric identity:
cos(2x) = 2cos^2(x) - 1
So, the equation can be rewritten as:
(2cos^2(2x) - 1) - (2cos^2(x) - 1) = 0
Simplifying this further:
2cos^2(2x) - 2cos^2(x) = 0
Divide both sides by 2:
cos^2(2x) - cos^2(x) = 0
Now, you can use another trigonometric identity:
cos(2x) = 2cos^2(x) - 1
To rewrite the equation again:
(2cos^2(x) - 1) - cos^2(x) = 0
Simplifying this:
cos^2(x) - 1 = 0
Now, it becomes a quadratic equation. Add 1 to both sides:
cos^2(x) = 1
Taking the square root of both sides:
cos(x) = ±1
Therefore, the possible solutions for x in the interval [0, 2pi) are:
x = 0, x = pi/2, x = pi, x = 3pi/2, x = 2pi
Please note that these solutions need to be checked back in the original equation to confirm if they are valid solutions.
To solve the equation cos4x - cos2x = 0, you've already made progress by rewriting it as cos(2(2x)) - cos(2x). Let's continue from there.
Using the identity cos(2θ) = 2cos^2(θ) - 1, you can rewrite the equation as:
2cos^2(2x) - 1 - 2cos^2(x) + 1 = 0
Simplifying further:
2cos^2(2x) - 2cos^2(x) = 0
Dividing both sides by 2:
cos^2(2x) - cos^2(x) = 0
Now, you can factor the left side using the difference of squares identity, which states that a^2 - b^2 = (a + b)(a - b):
(cos(2x) + cos(x))(cos(2x) - cos(x)) = 0
Now, set each factor equal to zero and solve for x:
1) cos(2x) + cos(x) = 0
To solve this equation, you can use the sum-to-product trigonometric identity, which states that cos(A) + cos(B) = 2cos((A + B)/2)cos((A - B)/2). Applying this identity:
2cos((2x + x)/2)cos((2x - x)/2) = 0
2cos(3x/2)cos(x/2) = 0
Since the product of the two factors is equal to zero, either cos(3x/2) = 0 or cos(x/2) = 0.
To solve each of these individually, you can find the values of x that make the cosine functions equal to zero in the specified interval [0, 2π).
2) cos(2x) - cos(x) = 0
Using the difference-to-product identity, which states that cos(A) - cos(B) = -2sin((A + B)/2)sin((A - B)/2), we can rewrite the equation:
-2sin((2x + x)/2)sin((2x - x)/2) = 0
-2sin(3x/2)sin(x/2) = 0
Again, since the product of the two factors is equal to zero, either sin(3x/2) = 0 or sin(x/2) = 0.
To solve each of these individually, you can find the values of x that make the sine functions equal to zero in the specified interval [0, 2π).
I hope this explanation helps you solve the equation and find all the solutions in the interval [0, 2π).