Solve 5sinx=-4cosx for 0 less than or equal to x <2pi
answers to nearest hunderth of a radian
To solve the equation 5sin(x) = -4cos(x), we'll use the identities sin(x) = cos(π/2 - x) and cos(x) = sin(π/2 -x).
First, let's rewrite the equation using these identities:
5sin(x) = -4cos(x)
5sin(x) = -4sin(π/2 - x)
Now, let's solve for x:
5sin(x) = -4sin(π/2 - x)
Divide both sides by 5:
sin(x) = (-4/5)sin(π/2 - x)
Now, we have two cases to consider:
Case 1: sin(x) = 0
If sin(x) = 0, then x can be any multiple of π.
Case 2: sin(x) ≠ 0
For sin(x) ≠ 0, we can cancel out sin(x) on both sides of the equation:
1 = (-4/5)cos(π/2 - x)
Now, we can solve for cos(π/2 - x):
cos(π/2 - x) = (5/4)
To find the values of x, we'll use the inverse cosine function:
π/2 - x = cos^(-1)(5/4)
Now, solve for x:
x = π/2 - cos^(-1)(5/4)
Note that the value of cos^(-1)(5/4) is undefined, which means there are no solutions for sin(x) ≠ 0.
In summary, the solutions to the equation 5sin(x) = -4cos(x) for 0 ≤ x < 2π to the nearest hundredth of a radian are:
1) x = 0 + 2πk, where k is an integer (for sin(x) = 0)
2) No solutions for sin(x) ≠ 0.