Question

Solve: 2sin(x) = 3^(1/2)tan(x)

So my teacher says in some cases we can't cancel the common factor or something :P so is this applicable here?? does this simplify to
cos(x) = (3^0.5)/2 ??
Cause if it does, then I don't need help :)

Answer 1

and @hartnn I spoke to my teacher: he said it's a typo xD *fist-in-the-air* ACHIEVEMENT UNLOCKED. My teacher admitted defeat. HA. I'm so happy :D

Answer 2

Is the right side, (3^1/2) times tangent, or is tangent also part of the exponent on the 3?

Answer 3

nopee, its times tan, tan's not part of the index.

Answer 4

ok I get ur question.
We couldn't cancel here too. There is sin (x) =0

Answer 5

Your solution looks correct then. Might want to still check your solutions, though, just in case.

Answer 6

i got \[\pm 30, \pm 330\]

Answer 7

^^ right, that would make 0 also a solution...

Answer 8

sin(0)=tan(0)

Answer 9

but i simplified it. Then x would not be 0.

Answer 10

Put x=0 into the original equation and check.

Answer 11

(and x=π . . .)

Answer 12

in the original equation it does work.

Answer 13

The original equation is what matters most.
The simplified forms are equivalent for some of the solutions, and can give more information, but they sometimes take away information too. The original equation is always boss.

Answer 14

\[2\sin(x)=\sqrt{3}\tan(x)\]
\[2\sin(x)-\sqrt{3}\tan(x)=0\]
\[2\sin(x)-\frac{ \sqrt{3}\sin(x) }{ \cos(x) }=0\]
\[\sin(x)(2-\frac{ \sqrt{3} }{ \cos(x) })=0\]\[\sin(x)=0 ......and .....2-\frac{ \sqrt{3} }{ \cos(x) }=0\]Now solve for x.

Answer 15

\[0,\pi/6,11\pi/6,\pi,2\pi\]are the answer for \[0<x<2\pi\]

Answer 16

nopess, answer is for -360 - 360 inclusive.

Answer 17

omg guys you know what, you've been really helpful, but i've got a few more questions to do, and I REALLY don't feel like changing it now. so i'll leave it as cos(x), and let the teacher mark me wrong :P BUT THANK YOU SO MUCHHH .

Answer 18

A similar thing happens with rational equations. If you cancel common factors top-and-bottom, you can lose information about one of the discontinuities.

e.g.
\[\large y=\frac{x^2+3x+2}{x^2-4} \rightarrow y=\frac{(x+2)(x+1)}{(x+2)(x-2)} \]
\[\large \rightarrow y=\frac{\cancel{(x+2)}(x+1)}{\cancel{(x+2)}(x-2)} \rightarrow y=\frac{(x+1)}{(x-2)}\]
If you just go by the final equation, you'll miss the x=/=-2 hole in the graph.

Answer 19

Just add the - ve sign for what I posted before and include them too.
Thar will be the answers.