trig help

izadapolak said:
i need help solving this trig problem. been stumped on it forever

tan = 2sinx
Click to expand...
You should be if we don't know what the tan is of.

If it's tan(x): you can move the 2sin(x) to the LHS, factorise and solve.
 
LHS = left-hand side.

Just get them on the same side to factorise.
 
tan(x) = 2sin(x)

becomes

tan(x) - 2sin(x) = 0

becomes

sin(x)/cos(x) - 2sin(x) = 0

Ok? You can factor out sin(x) from here.

Any requests to do the rest, without any effort of your own, will be ignored.
 
i cant factor out the sinx from here because it is over the cosx. that is not factorable because its one term. ive already tried chnaging it to sin and cos, squaring both sides and trying to put in trig identities and im getting no where
 
izadapolak said:
i cant factor out the sinx from here because it is over the cosx.
Click to expand...
\(\displaystyle \L\mbox{ \frac{\sin{x}}{\cos{x}} = \sin{x}\left(\frac{1}{\cos{x}}\right)}\)
 
You're making this problem far more difficult than it needs to be.

Maybe try this approach if you want to.

triangle25ub.gif


\(\displaystyle tan(x)=\frac{a}{b}\). Right?.

\(\displaystyle sin(x)=\frac{a}{c}\). OK?.

So we have \(\displaystyle \frac{a}{b}=\frac{2a}{c}\)

Now, get the a's on one side and cancel. You should end up with a fraction on one side and a trig ratio on the other(see triangle). An answer should be easily found from there.
 
Your approach is equivalent to saying

sin(x)/cos(x) = 2sin(x)

The sin(x)'s cancel

1/cos(x) = 2

See the flaw?
 
If we have:

\(\displaystyle \frac{\frac{a}{c}}{\frac{b}{c}}=2\frac{a}{c}\)

\(\displaystyle \frac{c}{b}=2\)

\(\displaystyle sec(x)=2\)

\(\displaystyle sec^{-1}(2)=\frac{{\pi}}{3}\)

The same as:

\(\displaystyle \frac{a}{b}=\frac{2a}{c}\)

\(\displaystyle ac=2ab\)

\(\displaystyle \frac{\sout{a}}{2\sout{a}}=\frac{b}{c}\)

Since \(\displaystyle \frac{b}{c}=cos(x)\)

\(\displaystyle \frac{1}{2}=cos(x)\)

\(\displaystyle cos^{-1}(\frac{1}{2})=x\)

\(\displaystyle x=\frac{{\pi}}{3}\)
 
Izadapolak, I usually do not jump into such an ongoing thread.

However, it seems that either you should be given the answer or nothing.

\(\displaystyle \L
\tan (x) = 2\sin (x)\quad \Rightarrow \quad \sin (x)\left( {\frac{1}{{\cos (x)}} - 2} \right) = 0\)

\(\displaystyle \L
\sin (x) = 0\quad \Rightarrow \quad x = k\pi\)

\(\displaystyle \L
\left( {\frac{1}{{\cos (x)}} - 2} \right) = 0\quad \Rightarrow \quad x = \frac{{ \pm \pi }}{3} + 2k\pi\)
 
Unco said:
What I am getting at: what if a=0?
Click to expand...

If the toad had wings he wouldn't bump his butt a hoppin'. :lol:

Sorry Unco, forgot about zero.


I was originally just going to do:

\(\displaystyle \frac{sin(x)}{cos(x)}=2sin(x)\)

\(\displaystyle \frac{1}{cos(x)}=2\)

\(\displaystyle x=sec^{-1}(2)=\frac{{\pi}}{3}\)

I suppose this has that darn zero element too. dagnubbit :oops:
 
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