# Need Help solving trigonometric equations. Im on the last problem and cant figure it out. Q: tanx=sinx?

so far i have:

tanx-sinx=0

sinx/cosx - sinx =0 because tanx= sinx/cosx

then

sinx-sinxcosx/cosx =0 after subtracting but i cant figure it out after this

### 8 Answers

- Jim BurnellLv 61 decade agoFavorite Answer
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)

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- PuggyLv 71 decade ago
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}

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- sahsjingLv 71 decade ago
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

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- emrichLv 44 years ago
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°

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- MathTutorLv 61 decade ago
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

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- yupchageeLv 71 decade ago
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

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- 1 decade ago
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

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- 1 decade ago
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)

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