# Thread: Solving trig equations- Is this right?

1. ## Solving trig equations- Is this right?

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

$\displaystyle 2sinAcosA-6sinA=0$

$\displaystyle 2sinA(cosA-3) = 0$

$\displaystyle sinA= 2$ or $\displaystyle cosA=3$

So, are there no solutions?

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

$\displaystyle 2sinAcosA-6sinA=0$

$\displaystyle 2sinA(cosA-3) = 0$

$\displaystyle sinA= 2$ or $\displaystyle cosA=3$

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