precalc

Solve sinx+cosx=0

Thanks!

  1. 👍
  2. 👎
  3. 👁
  1. sinx=-cosx
    tanx=-1
    x= arctan(-1)

    1. 👍
    2. 👎
    👤
    bobpursley
  2. We're not supposed to find it using arctan. We have to get one side to be 2 factors that equal zero when multiplied.

    1. 👍
    2. 👎

Respond to this Question

First Name

Your Response

Similar Questions

  1. Trigonometry

    4. Find the exact value for sin(x+y) if sinx=-4/5 and cos y = 15/17. Angles x and y are in the fourth quadrant. 5. Find the exact value for cos 165degrees using the half-angle identity. 1. Solve: 2 cos^2x - 3 cosx + 1 = 0 for 0

  2. Trig.......

    I need to prove that the following is true. Thanks (2tanx /1-tan^x)+(1/2cos^2x-1)= (cosx+sinx)/(cosx - sinx) and thanks ........... check your typing. I tried 30º, the two sides are not equal, they differ by 1 oh , thank you Mr

  3. Math

    How do I solve this? tan^2x= 2tanxsinx My work so far: tan^2x - 2tanxsinx=0 tanx(tanx - 2sinx)=0 Then the solutions are: TanX=0 and sinX/cosX = 2 sin X Divide through by sinX: we have to check this later to see if allowed (ie sinX

  4. Maths

    If is a n acute angle and tanx=3 4 evaluate cosx-sinx cosx+sinx

  1. Precalculus/Trig

    I can't seem to prove these trig identities and would really appreciate help: 1. cosx + 1/sin^3x = cscx/1 - cosx I changed the 1: cosx/sin^3x + sin^3x/sin^3x = cscx/1-cosx Simplified: cosx + sin^3x/sin^3x = cscx/1-cosx I don't

  2. Math

    (sinx - cosx)(sinx + cosx) = 2sin^2x -1 I need some tips on trigonometric identities. Why shouldn't I just turn (sinx + cosx) into 1 and would it still have the same identity?

  3. Pre-Calc

    Establish the identity. sinx + cosx/sinx - cosx = 1+2sinxcosx/2sin^2x-1

  4. Calculus

    Integrate sinx(cosx)^2dx using the substitution u=sinx. I know how to do this using u =cosx, but not sinx. The next problem on the homework was the same question except it asked to use u=cosx, so there couldn't have been a

  1. Math help again

    cos(3π/4+x) + sin (3π/4 -x) = 0 = cos(3π/4)cosx + sin(3π/4)sinx + sin(3π/4)cosx - cos(3π/4)sinx = -1/sqrt2cosx + 1/sqrt2sinx + 1/sqrt2cosx - (-1/sqrt2sinx) I canceled out -1/sqrt2cosx and 1/sqrt2cosx Now I have 1/sqrt sinx +

  2. maths - trigonometry

    I've asked about this same question before, and someone gave me the way to finish, which I understand to some extent. I need help figuring out what they did in the second step though. How they got to the third step from the

  3. Maths

    Solve this equation fo rx in the interval 0

  4. Calculus

    determine the absolute extreme values of the function f(x)=sinx-cosx+6 on the interval 0

You can view more similar questions or ask a new question.