To solve the equation 2cos(2x)cos(x) – 1 = 0, we first isolate the cosine terms:
2cos(2x)cos(x) = 1
Next, we can rewrite cos(2x) using the double angle formula:
cos(2x) = 2cos²(x) – 1
Substituting this back into our equation gives:
2(2cos²(x) – 1)cos(x) = 1
Expanding this, we have:
4cos²(x)cos(x) – 2cos(x) – 1 = 0
This is a cubic equation in terms of cos(x). We can denote y = cos(x). Then, our equation becomes:
4y³ – 2y – 1 = 0
Now, we can use numerical methods or graphing techniques to find the values of y. However, we can also try to find rational roots.
Using the Rational Root Theorem, we test possible values:
After testing, we find that y = 1/2 is a root. We can factor the polynomial as:
(y – 1/2)(4y² + 2y + 2) = 0
Now, we can set each factor to 0:
y – 1/2 = 0 gives y = 1/2.
For the quadratic 4y² + 2y + 2 = 0, we can check the discriminant:
D = b² – 4ac = 2² – 4(4)(2) = 4 – 32 = -28
This indicates there are no real solutions for this quadratic equation.
Thus, the only real solution is y = 1/2, which means cos(x) = 1/2.
The general solutions for cos(x) = 1/2 are:
x = rac{C0}{3} + 2k rac{C0}{3} ext{ and } x = rac{5C0}{3} + 2k rac{C0}{3}, ext{ where } k ext{ is any integer.}
Now, we need to find solutions within the interval [0, 2π]:
The two solutions are:
- x = rac{C0}{3} ext{ (approximately 1.05)}
- x = rac{5C0}{3} ext{ (approximately 5.24)}
Therefore, the final solutions to the given equation in the specified interval are:
- x = rac{C0}{3}
- x = rac{5C0}{3}