For the trig equation: 2sin(x)tan(x) = tan(x),
Why can't I just divide both sides by tan(x) and be left to solve only 2sin(x)=1?
Ahh, but then you would lose solutions....
2sin(x)tan(x) = tan(x)
2sinx tanx - tanx = 0
tanx(2sinx - 1) = 0
tanx = 0 or sinx = 1/2
so tanx = 0
x = 0, 180
or
sinx = 1/3
x = 30° , 150°
so x = 0°, 30°, 150°, 180°, 360° in the domain 0 ≤ x ≤ 360°
I do not see why not.
x = 30 degrees or 180 - 30 etc
Okay thanks! Makes sense now
When solving trigonometric equations, it is important to be aware of the domain and range of the functions involved, as well as any potential solutions that may arise from them. In the given equation, 2sin(x)tan(x) = tan(x), dividing both sides by tan(x) seems like a good initial step to isolate the variable.
However, dividing both sides by tan(x) can lead to a loss of information about the possible solutions. This is because the function tan(x) has a domain where it is not defined or where it becomes zero. Specifically, tan(x) is undefined when x is equal to (2n + 1)π/2, where n is an integer. Additionally, tan(x) becomes zero when x is equal to nπ, where n is an integer.
If you divide both sides by tan(x), it means you are assuming that tan(x) is not zero. While this is true for some values of x, it is not true for the entire domain of tan(x). Therefore, when dividing both sides of the equation by tan(x), you might eliminate potential solutions that correspond to the values of x where tan(x) is zero.
To handle this situation properly, you need to consider both cases: when tan(x) is zero, and when tan(x) is not zero. By doing this, you will ensure that you explore all possible solutions for the equation.
In summary, while dividing both sides by tan(x) seems like a convenient move, it can lead to the loss of potential solutions. To solve the equation 2sin(x)tan(x) = tan(x) correctly, you should consider both cases separately: when tan(x) is zero and when it is not zero.