Can someone please explain on how to solve this equation so I can do the others that are similar to this one?
Find the solutions of the equations that are in the interval [0,2pi)
cos x cot^2 x= cos x
Thank you!
cos x cot^2 x= cos x
cos x cot^2 x - cos x = 0
cos x (cot^2 x - 1) = 0
cosx = 0
cotx = ±1
Now you can just find x, right?
To solve the equation cos x cot^2 x = cos x, we can follow these steps:
Step 1: Simplify the equation if possible
Start by dividing both sides of the equation by cos x:
(cot^2 x) = 1
Step 2: Rewrite cot^2 x in terms of sin x and cos x
Recall that cot x = cos x / sin x. Therefore, cot^2 x = (cos^2 x) / (sin^2 x).
Substituting this back into the equation, we have:
(cos^2 x) / (sin^2 x) = 1
Step 3: Solve for cos x and sin x
We know that cos^2 x + sin^2 x = 1, so we can rewrite the equation as:
(cos^2 x) / (sin^2 x) = cos^2 x + sin^2 x
Now, let's simplify this equation:
(cos^2 x) / (sin^2 x) - (cos^2 x + sin^2 x) = 0
Multiplying both sides by sin^2 x, we get:
cos^2 x - cos^2 x sin^2 x - sin^4 x = 0
Step 4: Factor and solve
Let's factor out cos^2 x:
cos^2 x (1 - sin^2 x) - sin^4 x = 0
Using the Pythagorean Identity, sin^2 x = 1 - cos^2 x, we can simplify further:
cos^2 x (cos^2 x) - sin^4 x = 0
(cos^4 x) - sin^4 x = 0
Now, we have a difference of squares, which can be factored into:
(cos^2 x - sin^2 x) (cos^2 x + sin^2 x) = 0
Since cos^2 x + sin^2 x = 1 (from the Pythagorean Identity), the equation becomes:
(cos^2 x - sin^2 x) (1) = 0
(cos^2 x - sin^2 x) = 0
Now, we have two separate equations to solve:
1) cos^2 x - sin^2 x = 0
2) 1 = 0 (which is not possible)
For equation 1), we can use the difference of squares formula, cos^2 x - sin^2 x = (cos x - sin x)(cos x + sin x) = 0.
Setting each factor equal to zero, we have two possible solutions:
cos x - sin x = 0 OR cos x + sin x = 0
For equation cos x - sin x = 0, we rearrange to get:
cos x = sin x
Now, we divide both sides by cos x (since it cannot be zero in the given interval [0, 2pi)):
1 = tan x
This tells us that x is an angle for which the tangent is equal to 1. In the given interval [0, 2pi), the solutions for this equation are x = pi/4 and x = 5pi/4.
For equation cos x + sin x = 0, we rearrange to get:
cos x = -sin x
Dividing both sides by cos x (again, noting that cos x cannot be zero in the given interval):
1 = -tan x
This tells us that x is an angle for which the tangent is equal to -1. In the given interval [0, 2pi), the solution for this equation is x = 3pi/4.
So, the solutions for the original equation cos x cot^2 x = cos x in the interval [0, 2pi) are x = pi/4, x = 3pi/4, and x = 5pi/4.