Equation $2\cos(x)-3\tan(x)=0$

Question

I solved this equation $2\cos(x)-3\tan(x)=0$ and I got, $\frac{1}{2}=\sin(x)$ and $-2=\sin(x)$.

For the first solution I got $\arcsin(1/2)=x, 30°=x$, but second is invalid because the domain of arcsin can be only between $-1$ and $1$.

Right???

Thanks.

EDIT:

My question: is the solution valid for $\arcsin(-2)$, because the domain of $\arcsin()$ is $-1<x<1$

amWhy · Accepted Answer · 2014-02-08T15:44:33.827

3

You're entirely correct. Good work! (We throw out the solution $\sin x = -2$ for precisely the reason you give. I.e., There is no $x$ such that $\sin x = -2$.)

edited Feb 08 '14 at 15:44

answered Feb 08 '14 at 15:39

amWhy

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this answer could be safely fit into a comment, right?? – lab bhattacharjee Feb 08 '14 at 15:41
1

@lab Sure it could. But it puts closure on the question, so as not to be "unanswered." What does answering as a CW hurt? – amWhy Feb 08 '14 at 15:42
Surely this a nice answer since it's useful for the OP! – Feb 09 '14 at 12:57

Equation $2\cos(x)-3\tan(x)=0$

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