1. Simple trig equation

5sin(A)cos(A) - sin(A) = 0 (A is between 0 and 360 degrees)

by dividing through by sin(A) I get:

cos(A) = 1/5

so A = 78.5, 281.5

Though clearly A = 0 also satisfies this equation so could somebody please explain why I have missed it in my method?

2. Re: Simple trig equation

by dividing, you "lost" solutions. factoring is the way to go here ...

$5\sin(A)\cos(A) - \sin(A) = 0$

$\sin(A)[5\cos(A) - 1] = 0$

$\sin(A) = 0 \implies A = 0 \, , \, 180 \, , \, 360$

$\cos(A) = \frac{1}{5} \implies A = \arccos\left(\frac{1}{5}\right) \, , \, 360 - \arccos\left(\frac{1}{5}\right)$

3. Re: Simple trig equation

Thanks but I still don't understand how those solutions were lost? If attempting a question like this how will I know that dividing will lose solutions? Are you able to explain this in a way that I can understand?

Thanks

4. Re: Simple trig equation

Has no one ever explained to you that you cannot divide by 0? Either sin(A) is not 0 so you can divide by it or sin(A)= 0.

A better way of doing this problem would be to write 5 sin(A)cos(A)- sin(A)= sin(A)(5 cos(A)- 1)= 0. The product of two numbers is 0 if and only if one or the other of them is 0: either sin(A)= 0 or 5cos(A)- 1= 0.

5. Re: Simple trig equation

Hello, kinhew93!

Thanks, but I still don't understand how those solutions were lost.

You will lose a solution if you divide by the variable
. . (or an expression containing the variable).

Suppose we have the equation: . $x^2 - 3x \:=\:0$ .[1]

. . and you know it has two solutions, right?

The proper procedure is:

. . Factor: . $x(x-3) \:=\:0$
. . Set each factor equal to zero and solve: . $\begin{Bmatrix}\boxed{x \:=\:0} \\ x-3\:=\:0 & \Rightarrow& \boxed{x \:=\:3} \end{Bmatrix}$

But suppose you decide to divide [1] by $x\!:\;\;\frac{x^2}{x} - \frac{3x}{x} \:=\:\frac{0}{x}$
Then you have: . $x - 3 \:=\:0$
. . which has only one solution: . $x = 3$

See?

6. Re: Simple trig equation

Seems so obvious now... thanks