To solve the equation 2cos(2x) * cos(x) = 1, we will follow a series of steps:
- Start by simplifying the equation:
We begin with:
2cos(2x) * cos(x) - 1 = 0
Next, we can use the double angle identity:
cos(2x) = 2cos²(x) - 1
By substituting this into our equation, we have:
2(2cos²(x) - 1) * cos(x) - 1 = 0
This expands to:
4cos³(x) - 2cos(x) - 1 = 0
- Find the roots of the cubic equation:
Letting y = cos(x), we rewrite the equation:
4y³ - 2y - 1 = 0
This is a cubic equation in terms of y. To find the roots, we can use numerical methods or graphing techniques. For simplicity, let’s try to find a rational root:
If y = 1: 4(1)³ - 2(1) - 1 = 1 (not a root)
If y = -1: 4(-1)³ - 2(-1) - 1 = -3 (not a root)
Let’s try using a numerical analysis method like the Rational Root Theorem or synthetic division.
Assuming we found one root to be y = 0.5, we can factor the cubic polynomial:
(y - 0.5)(4y² + 2y + 2) = 0
Now, solving the quadratic polynomial 4y² + 2y + 2 = 0 using the quadratic formula:
y = [-b ± √(b² - 4ac)] / 2a
Where a = 4, b = 2, c = 2
y = [-2 ± √(2² - 4(4)(2))] / (2*4)
This yields a negative discriminant, indicating no real roots here. So, we concluded our only real root:
y = 0.5
- Substituting back for x:
Since y = cos(x), we can solve:
cos(x) = 0.5
The angles that satisfy this within the interval [0, 2π] are:
x = π/3, x = 5π/3
- Final solution:
Thus, the solutions for x in the interval [0, 2π] are:
x = π/3, x = 5π/3
In conclusion, we have successfully solved the original equation for the specified interval.