Answer: To solve the equation cos^2(x) - sin^2(x) = sin(x) over the given domain [0, 2π), we'll first use trigonometric identities to simplify the equation and then solve for x.
Step 1: Apply the Pythagorean identity: cos^2(x) - sin^2(x) = 1 - sin^2(x) - sin^2(x)
Step 2: Combine like terms: 1 - 2sin^2(x) = sin(x)
Step 3: Move all terms to one side to set the equation to zero: 2sin^2(x) + sin(x) - 1 = 0
Now, we have a quadratic equation in terms of sin(x). To solve this, we can use factoring or the quadratic formula:
Let sin(x) = t
2t^2 + t - 1 = 0
Step 4: Factor the quadratic equation or use the quadratic formula to find the solutions for t (sin(x)).
The quadratic doesn't factor easily, so we'll use the quadratic formula:
t = [-b ± √(b^2 - 4ac)] / 2a
where a = 2, b = 1, and c = -1.
t = [-1 ± √(1 - 4(2)(-1))] / 2(2)
t = [-1 ± √(1 + 8)] / 4
t = [-1 ± √9] / 4
Now, find the two possible values for t:
t = (-1 + 3) / 4 = 2/4 = 1/2
t = (-1 - 3) / 4 = -4/4 = -1
Step 5: Convert the values of t back to values of sin(x).
Since sin(x) lies between -1 and 1, we can reject the value of t = -1 (it is outside the domain of sin(x)).
Therefore, t = sin(x) = 1/2
Step 6: Find the values of x corresponding to sin(x) = 1/2 in the given domain [0, 2π).
sin(x) = 1/2 corresponds to two values of x in the domain [0, 2π):
x = π/6 (30 degrees)
x = 5π/6 (150 degrees)
Step 7: Final solution.
The solutions for the equation cos^2(x) - sin^2(x) = sin(x) over the domain [0, 2π) are x = π/6 and x = 5π/6.
Why must the domain be restricted?
The domain is restricted to [0, 2π) because the original equation involves trigonometric functions (cosine and sine). Trigonometric functions are periodic, meaning they repeat their values at regular intervals. The interval [0, 2π) represents one full cycle of the sine and cosine functions, and we want to find solutions within this range.
How is this different from solving a polynomial equation?
Solving a polynomial equation involves finding the values of the variable that make the polynomial equation true, without any domain restrictions. Polynomial equations can have real solutions, complex solutions, or no solutions at all. However, when dealing with trigonometric equations, we often need to restrict the domain to find solutions that lie within a specific range due to the periodic nature of trigonometric functions. Additionally, trigonometric equations can have infinitely many solutions over the entire real number line.