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!
I don't much like it, but you can still get rid of the cosine using $\displaystyle \cos(x)\;=\;\sqrt{1-\sin^{2}(x)}$.
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
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...
Freya meant that he already did what moolimanj suggested. He just multiplied by -1 in the end.I don't know what to say to the suggestion from moolimanj...
$\displaystyle \sin{x}\tan{x}+\tan{x}-2\sin{x}+\cos{x}=0$
$\displaystyle \frac{\sin^2{x}}{\cos{x}} + \frac{\sin{x}}{\cos{x}} - \frac{2\sin{x}\cos{x}}{cos{x}} + \frac{\cos^2{x}}{\cos{x}} = 0$
$\displaystyle \frac{\sin^2{x} + \sin{x} - 2\sin{x}\cos{x} + \cos^2{x}}{\cos{x}} = 0$
$\displaystyle \sin^2{x} + \sin{x} - 2\sin{x}\cos{x} + \cos^2{x} = 0$
$\displaystyle 1 + \sin{x} - 2\sin{x}\cos{x} = 0$
$\displaystyle 2\sin{x}\cos{x} - \sin{x} -1 = 0$