Math Help - Solving trig equations- Is this right?

1. Solving trig equations- Is this right?

$sin(2A)=6sin(A)$

$2sinAcosA-6sinA=0$

$2sinA(cosA-3) = 0$

$sinA= 2$ or $cosA=3$

So, are there no solutions?

2. Originally Posted by coasterguy
$sin(2A)=6sin(A)$

$2sinAcosA-6sinA=0$

$2sinA(cosA-3) = 0$

$sinA= 2$ or $cosA=3$

So, are there no solutions?
From the line $2\sin A\left(\cos A-3\right)=0$, you should get the two equations $2\sin A=0$ or $\cos A-3=0$. If $A\in\mathbb{R}$, the second one is never true (If $A\in\mathbb{C}$, then it would be a different story). The first one has a solution, since $2\sin A=0\implies \sin A=0$ which is true when $A=k\pi$, where $k\in\mathbb{Z}$ ( $\mathbb{Z}$ is the set of integers).

Does this make sense?

3. solve this equation

let z = { -1,-2,-3,....,0.1.2.3....}= the set of integre numbers