OK this is a complete re-do.
You had the right idea.
tanx - sinx = 0
sinx/cosx - sinx = 0
cosx(sinx/cosx) - cosx(sinx) = cosx(0)
sinx - sinx cosx = 0
Now you can factor:
sinx(1 - cosx) = 0
So either sinx = 0 (which is at 0, pi, 2pi, etc)
Or cosx = 1 (which is at 0, 2pi, 4i, etc)
Since sinx = 0 provides more solutions than cosx = 1, the answer is "everywhere that sinx = 0"
0 +/- nπ
(where n is any integer)
2006-12-28 06:36:56
·
answer #1
·
answered by Jim Burnell 6
·
1⤊
0⤋
You're on the right track.
tan(x) - sin(x) = 0
sinx/cosx - sinx = 0
Putting them under a common denominator,
sinx/cosx - sinxcosx/cosx = 0
(sinx - sinxcosx)/cosx = 0
At this point, we equate the numerator to 0
sinx - sinxcosx = 0. Factoring, we get
sin(x) [1 - cosx] = 0
And we obtain two equations equated to 0.
sin(x) = 0
1 - cos(x) = 0 (implying cos(x) = 1)
The solution set to sin(x) = 0, assuming a restricted domain of
0 <= x < 2pi, is x = 0, pi.
The solution set to cos(x) = 1 is x = 0
Therefore, x = {0, pi}
2006-12-28 07:17:55
·
answer #2
·
answered by Puggy 7
·
0⤊
0⤋
First, assume you only need to find the solutions on [ 0, 2pi).
Start from
tanx = sinx
Collect two terms in one side,
tanx - sinx = 0
Take the common factor sinx out,
sinx(1/cosx -1) = 0
Therefore, either sinx = 0, which gives solutions x = 0, pi, or cosx = 1 which gives only one solution x = 0.
Combine all solutions, x = 0 or pi.
If the domain is not fixed on [0,2pi), you can extend the solutions by adding periods ±n pi
2006-12-28 06:52:05
·
answer #3
·
answered by sahsjing 7
·
0⤊
0⤋
You`ve controlled to squeeze some EQUATIONS into an EQUATION ! question a million cos x (a million - 2 sin x) = 0 cos x = 0 , sin x = a million/2 x = ninety° , 270° , 30° , a hundred and fifty° question 2 ought to it somewhat is :- 2 sin ² x + 5x - 3 = 0 ? (2 sinx - a million)(sin x + 3) = 0 sin x = a million/2 x = 30° , a hundred and fifty° question 3 sin ² x = a million / 4 sin x = ± (a million/2) x = 30° , a hundred and fifty° , 210° , 330° question 4 cos x ( sin x - a million) = 0 cos x = 0 , sin x = a million x = ninety° , 270°
2016-10-19 02:29:12
·
answer #4
·
answered by Anonymous
·
0⤊
0⤋
sinx/cosx= sin x
One solution is sin x = 0, you have there any integer multiple of pi as a solution
If sin x is not 0 (if x is not a integer multiple from pi), then you can cancell
So, you have 1/cos x = 1, or 1 = cos x. But the only x who are solution from this equation are the multiple from 2pi. So, you don´t obtain new solutions.
Ana
2006-12-28 06:50:37
·
answer #5
·
answered by MathTutor 6
·
0⤊
0⤋
tanx=sinx tan=sin/cos
sin x/cos x=sin x divide both sides by sin x
1/cos x=1
cos x=1
x-arccos 1=0°, 360°, 720° etc
2006-12-28 07:08:35
·
answer #6
·
answered by yupchagee 7
·
0⤊
0⤋
taking it from what you have so far:
multiply both sides by cosx
sinx - sinxcosx=0
factor out sinx
sinx(1-cosx)=0
divide both sides by sinx
1-cosx=0
cosx = 1
2006-12-28 06:40:31
·
answer #7
·
answered by joycedomingo 3
·
0⤊
0⤋
tangent does not equal sine, only when x=0.
Proof:
Consider x=Pi/4
tan(Pi/4) = 1 while sin(Pi/4) = 1/sqrt(2)
2006-12-28 06:37:38
·
answer #8
·
answered by Tony O 2
·
0⤊
1⤋