Solving Trig Equation - Tangent

[Dudethereskelsey]

In this image, I attempted to solve a quadratic trig equation using two different methods. I numbered them 1 and 2. It has already been made clear to me that method 1 is correct and method 2 is incorrect. My question is this: why do the two methods yield totally different results? I do not see any errors in method 2, so it seems like it should yield equivalent results to method 1. Also, no common angle has a tangent of 1/2, which confuses me about using method 1. If neither method has any mistakes, why do they give different answers? Shouldn't everything be equivalent?

 

[willyengland]

For method 1 you set 2tan(x) - 1 = 0. This is allowed, because only if one of the two factors is Zero, the complete product becomes Zero.
For method 2 you set tan(x) = 1. This is not correct, because also other combinations would allow for 1 as a result, .e.g 0,5 * 2 = 1.

 

[Dr.Peterson]

View attachment 14471
In this image, I attempted to solve a quadratic trig equation using two different methods. I numbered them 1 and 2. It has already been made clear to me that method 1 is correct and method 2 is incorrect. My question is this: why do the two methods yield totally different results? I do not see any errors in method 2, so it seems like it should yield equivalent results to method 1. Also, no common angle has a tangent of 1/2, which confuses me about using method 1. If neither method has any mistakes, why do they give different answers? Shouldn't everything be equivalent?

First, as has already been said or implied, factoring is appropriate only when the other side is zero. Your method 2 amounts to solving x(2x+1) = 1 by assuming that either x or 2x+1 must equal 1, but that is false. There are many ways to obtain 1 as a product, and even the one you chose requires both factors to be 1. So this is a very serious mistake.

In particular, what you have shown is that the equation is true if one x is 45° and the other is 0; that is not what it means for x to be a solution of the equation!!!

Method 1 is the standard way to solve (2x-1)(x+1) = 0, using the fact that the only way a product can be 0 is if one (or both) of the factors are zero.

As to the solution of tan(x) = 1/2, solutions do not have to be "common angles". The solution is [MATH]x = \tan^{-1}(1/2) + 180n[/MATH], for any integer n. You can use a calculator to obtain the specific solutions in your interval.

 

[Steven G]

\(\displaystyle 1*5\neq 1, \, 1*23\neq 1, \, 28*1\neq 1,\, 21*1\neq 1.\) Only 1*1 = 1 (if one of the 2 numbers must be 1). The thing with 0 is that 0*23=0*4=23*0 = -25*0=0.

So if you have x*y =1 it does not follow that x=1. It can be that x=4/5 and y=5/4 so none of the two numbers are 1.