promotion image of download ymail app
Promoted

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

Relevance
  • 1 decade ago
    Favorite 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)

    • Commenter avatarLogin to reply the answers
  • Puggy
    Lv 7
    1 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}

    • Commenter avatarLogin to reply the answers
  • 1 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

    • Commenter avatarLogin to reply the answers
  • emrich
    Lv 4
    4 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°

    • Commenter avatarLogin to reply the answers
  • How do you think about the answers? You can sign in to vote the answer.
  • 1 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

    • Commenter avatarLogin to reply the answers
  • 1 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

    • Commenter avatarLogin to reply the answers
  • 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

    • Commenter avatarLogin to reply the answers
  • 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)

    • Commenter avatarLogin to reply the answers
Still have questions? Get your answers by asking now.