Find solutions on interval 0,2pie
(cosx/1+sinx) + (1+sinx/cosx) = -4
cosx/1+sinx or cosx/(1+sinx)
1+six/cosx or (1+sinx)/cosx ?
The second ones
Do you know how to solve equation
a*sinx+b*cosx=c ?
To find the solutions to the equation (cosx/1+sinx) + (1+sinx/cosx) = -4 on the interval 0 to 2π, we can follow these steps:
1. Combine the fractions on the left-hand side of the equation using a common denominator. The common denominator in this case is cosx(1 + sinx):
[cosx(1 + sinx) + (1 + sinx)] / (1 + sinx)(cosx) = -4
2. Simplify the numerator:
[cosx + cosx(sin^2(x))] / (1 + sinx)(cosx) = -4
3. Expand (sin^2(x)):
[cosx + cosx(1 - cos^2(x))] / (1 + sinx)(cosx) = -4
4. Distribute cosx:
[cosx + cosx - cosx(cos^2(x))] / (1 + sinx)(cosx) = -4
5. Simplify:
[2cosx - cosx(cos^2(x))] / (1 + sinx)(cosx) = -4
6. Multiply through by (1 + sinx)(cosx) to remove the denominators:
2cosx - cosx(cos^2(x)) = -4(1 + sinx)(cosx)
7. Expand and simplify:
2cosx - cosx(cos^2(x)) = -4cosx - 4cosx(sin(x))
8. Rearrange the equation:
2cosx + 4cosx + cosx(cos^2(x)) + 4cosx(sin(x)) = 0
9. Combine like terms:
10cosx + cos^3(x) + 4sin(x)cosx = 0
Now we have an equation that we can solve for the values of x on the interval 0 to 2π. This is a transcendental equation, which means there is no general algebraic method to find its solutions. However, we can use numerical methods or a graphing calculator to approximate the solutions.