C(1)=6
c(2)=c(2-1)-16=c(1)-16=6-16=-10
c(3)=c(3-1)-16=c(2)-16=-10-16=-26
Cos x + 2sec x = -3
cos x + 2/cos x = -3
cos^2 x + 2 = -3cos x
cos^2 x + 3cos x + 2 = 0
(cos x + 1)(cos x + 2) = 0
cos x + 1 = 0 or cos x + 2 = 0
cos x = -1 or cos x = -2 [but cos x cannot be -2, so we discard it]
x = arccos(-1)
x = (2n + 1)π
Answer:
The solutions are and .
Step-by-step explanation:
We have the following equation:
.
The first step to solve this problem is using
We replace in the equation 1, find the values of y, and then we replace in equation 2) to find the values of x.
To solve the equations, it is important to know how we find the roots of a second order polynomial.
Given a second order polynomial expressed by the following equation:
This polynomial has roots such that , given by the following formulas:
In this problem, we have
So
So:
The values of y are
We also have that:
So
And
There is no real solution for this. So our only solutions are and .
Hope this will help u...:)
Answer:
answer is 0.33333333
Step-by-step explanation: