Solve the equation (x+3)(x-3) = -9. What are the solutions to the equation if you replace x first with sinx and then with cosx, where 0≤x≤2pi?
x^2 - 9 = -9 ??
x^2 = 0
sin^2 x = 0
x = 0 or 180 degrees
cos^2 x = 0
x = 90 or 270 degrees
To solve the equation (x+3)(x-3) = -9, we can expand the left side of the equation using the difference of squares formula:
(x+3)(x-3) = x^2 - 9.
Now, let's set the expanded equation equal to -9:
x^2 - 9 = -9.
Adding 9 to both sides, we get:
x^2 = 0.
Therefore, the solutions to the equation (x+3)(x-3) = -9 are x = 0.
To find the solutions to the equation using sin(x) and cos(x), let's substitute x with sin(x) first.
(sin(x))^2 = 0.
Taking the square root of both sides, we get:
sin(x) = 0.
The solutions to sin(x) = 0 between 0≤x≤2pi are x = 0 and x = pi.
Now, let's substitute x with cos(x) and solve for the equation:
(cos(x))^2 = 0.
Taking the square root of both sides, we get:
cos(x) = 0.
The solutions to cos(x) = 0 between 0≤x≤2pi are x = 0.5*pi and x = 1.5*pi.
Therefore, when x is replaced with sin(x), the solutions are x = 0 and x = pi, and when x is replaced with cos(x), the solutions are x = 0.5*pi and x = 1.5*pi within the range 0≤x≤2pi.
To solve the equation (x+3)(x-3) = -9, we can follow these steps:
1. Expand the equation: (x+3)(x-3) = x² - 9.
2. Set the equation equal to -9: x² - 9 = -9.
3. Add 9 to both sides: x² = 0.
4. Take the square root of both sides: x = ±√0.
5. Simplify: x = 0.
Therefore, the solution to the equation for x is x = 0.
Now, let's replace x with sinx and solve the equation:
1. Substitute sinx into the equation: (sinx + 3)(sinx - 3) = -9.
2. Expand the equation: sin²x - 9 = -9.
3. Add 9 to both sides: sin²x = 0.
4. Take the square root of both sides: sinx = ±√0.
5. Simplify: sinx = 0.
Since sinx = 0 when x = 0, π, and 2π, the solutions to the equation for sinx are x = 0, π, and 2π.
Next, let's replace x with cosx and solve the equation:
1. Substitute cosx into the equation: (cosx + 3)(cosx - 3) = -9.
2. Expand the equation: cos²x - 9 = -9.
3. Add 9 to both sides: cos²x = 0.
4. Take the square root of both sides: cosx = ±√0.
5. Simplify: cosx = 0.
Since cosx = 0 when x = π/2 and 3π/2, the solutions to the equation for cosx are x = π/2 and 3π/2.
In summary:
- For the original equation, the solution is x = 0.
- For the equation with sinx, the solutions are x = 0, π, and 2π.
- For the equation with cosx, the solutions are x = π/2 and 3π/2.