Math Help - Simple trig equation...

1. Simple trig equation...

Can someone help with this (solve for x):

sinxtanx+tanx-2sinx+cosx=0

I end up with:
2sinxcosx-sinx-1=0 but can't get further....

please help!

2. I don't much like it, but you can still get rid of the cosine using $\cos(x)\;=\;\sqrt{1-\sin^{2}(x)}$.

Rearrange, Square, Rearrange, and notice that it factors a little, (sin(x)+1)(Other Stuff) = 0.

Irritatingly, sin(x) + 1 = 0 leads to (3/2)pi which is NOT a solution of the original expression. Your task is to find the remaining Real solution to the cubic expression in sin(x). I could tell you what it is, exactly, but you totally do NOT want to know that. Perhaps a numerical solution would be best. Start looking around x = 3.5

3. Note also that tan(x) can be written as Sin(x)/cos(x)

4. Thanks TKHunny but I don't like it either. sinx = -1 is not in the domain set for the problem as tanx = oo when sinx = -1.

Trying your suggestion I got the cubic:

4sin^3x - 4Sin^2x + sinx +1 = 0

which has a remainder of -8 when subbing for sinx = -1, i.e it is not a factor as it becomes:

(sinx + 1)(4sin^2 - 8sinx + 9) - 8 = 0

Graphing seems the only way to solve this and it gives a solution of x = 3.4968 (or sinx = -0.3478).

I think the problem is messed up as this is way too difficult for grade math.

I don't know what to say to the suggestion from moolimanj...

5. Originally Posted by Air
$\sin x\tan x+\tan x-2\sin x+\cos x=0$

$\sin x \left(\frac{\sin x}{\cos x}\right) +\left(\frac{\sin x}{\cos x}\right)-2\sin x+\cos x=0$

Multiplying through by $\cos x$

$\sin ^2 x + \sin x - 2\sin x \cos x + \cos ^2x=0$

...Continue and see if this would work.
I don't know what to say to the suggestion from moolimanj...
Freya meant that he already did what moolimanj suggested. He just multiplied by -1 in the end.

$\sin{x}\tan{x}+\tan{x}-2\sin{x}+\cos{x}=0$

$\frac{\sin^2{x}}{\cos{x}} + \frac{\sin{x}}{\cos{x}} - \frac{2\sin{x}\cos{x}}{cos{x}} + \frac{\cos^2{x}}{\cos{x}} = 0$

$\frac{\sin^2{x} + \sin{x} - 2\sin{x}\cos{x} + \cos^2{x}}{\cos{x}} = 0$

$\sin^2{x} + \sin{x} - 2\sin{x}\cos{x} + \cos^2{x} = 0$

$1 + \sin{x} - 2\sin{x}\cos{x} = 0$

$2\sin{x}\cos{x} - \sin{x} -1 = 0$