cosx sinx -sinx=0
i solve it and get 3 answer but in the back of the book, it ony gave 2 answer.
cosx sinx -sinx=0
sinx(cosx - 1) = 0
sinx = 0 or cosx = 1
within 0 ≤ x ≤ 360°
from sinx = 0 , x = 0 , 180° , 360°
from cosx = 1 , x = 0 , 360°
so x = 0 , 180°, 360°
Usually both 0 and 360 are not in the domain, so the book probably only lists 0 and 180 as solutions.
To solve the equation cos(x) sin(x) - sin(x) = 0, we can first factor out common terms. Notice that both cos(x) sin(x) and -sin(x) have sin(x) as a common factor.
So, let's factor out sin(x) from the equation:
sin(x) * (cos(x) - 1) = 0
Now, we have a product equal to zero. According to the Zero Product Property, if a * b = 0, then either a = 0 or b = 0.
In our equation, sin(x) * (cos(x) - 1) = 0, we have two possibilities:
1. sin(x) = 0
2. (cos(x) - 1) = 0
For the first possibility, sin(x) = 0, we know that sin(x) is equal to 0 when x is an integer multiple of π (pi). So, one solution is x = nπ, where n is an integer.
For the second possibility, cos(x) - 1 = 0, we can solve for cos(x):
cos(x) = 1
In other words, x can be any angle whose cosine is equal to 1. The only angle where this is true is x = 0.
Therefore, the solutions to the equation cos(x) sin(x) - sin(x) = 0 are:
- x = nπ, where n is an integer, and
- x = 0.