trig

posted by Alex

2sin(x)cos(x)+cos(x)=0

I'm looking for exact value solutions in [0, 3π]

So I need to find general solutions to solve the equation. But do I eliminate cos(x), like this...

2sin(x)cos(x)+cos(x)=0

2sin(x)cos(x)= -cos(x)

2sin(x) = -1

sin(x) = -1/2 at 4pi/3 and 5pi/3

and then use those general solutions to find my exact values? Or do I need to incorporate the cos(x) somehow, by factoring it out of the original equation like...

2sin(x)cos(x)+cos(x)=0

cos(x)[2sin(x)+1]=0

and then use general solutions for that?

  1. Damon

    when you divide both sides by cos x, you divide by zero when cos x = 0. Not allowed.
    However your equation is satisfied if cos x = 0, so at x = pi/2 and at x = 3 pi/2
    Your other reasoning is also valid
    sin x = -1/2 but I see those as x = pi+pi/6 = 7 pi/6 and at pi-pi/6 = 5 pi/6

  2. Alex

    Not dividing both sides by cos x, subtracting cos x from the left side first, then dividing by cos x. Would that work?

  3. Damon

    What you did later worked fine but dividing by cos x, before or after moving it to the right, is suspect in my mind.
    However what you did later:
    2sin(x)cos(x)+cos(x)=0
    then
    cos(x)[2sin(x)+1]=0
    so zero when cos x = 0
    or when sin x = -1/2
    is just fine and safe to my mind.
    I disagee with you about where in quadrants 3 and 4 sin x = -1/2

  4. Damon

    sin x = -1/2 but I see those as x = pi+pi/6 = 7 pi/6 and at 2pi-pi/6 = 11pi/6

  5. Damon

    cos(x)[2sin(x)+1]=0
    so zero when cos x = 0
    or when sin x = -1/2
    note
    this is just like factoring a quadratic
    x^2 - 4x + 3 = 0
    (x-3) (x-1) = 0
    satisfied when x-3 = 0 , so x = 3
    and when x-1 = 0 so x = 1

  6. Alex

    My mistake about the locations of where sin(x) is equal to -1/2... &#*@ now I have to redo two lengthy problems.

    Thanks, though.

Respond to this Question

First Name

Your Answer

Similar Questions

  1. tigonometry

    expres the following as sums and differences of sines or cosines cos8t * sin2t sin(a+b) = sin(a)cos(b) + cos(a)sin(b) replacing by by -b and using that cos(-b)= cos(b) sin(-b)= -sin(b) gives: sin(a-b) = sin(a)cos(b) - cos(a)sin(b) …
  2. math (trig)

    Find sin(x/2) if sin(x)= -0.4 and 3pi/2 < or equal to (x) < or equal to 2pi Let's use cos 2A = 1 - 2sin 2 A and we can match cos x = 1 - 2sin 2 (x/2) so we will need cos x we know sin x = -.4 and x is in the fourth quadrant, …
  3. Math - Solving Trig Equations

    What am I doing wrong? Equation: sin2x = 2cos2x Answers: 90 and 270 .... My Work: 2sin(x)cos(x) = 2cos(2x) sin(x) cos(x) = cos(2x) sin(x) cos(x) = 2cos^2(x) - 1 cos(x) (+/-)\sqrt{1 - cos^2(x)} = 2cos^2(x) - 1 cos^2(x)(1 - cos^2(x))
  4. Precal

    I do not understand how to do this problem ((sin^3 A + cos^3 A)/(sin A + cos A) ) = 1 - sin A cos A note that all the trig terms are closed right after there A's example sin A cos A = sin (A) cos (A) I wrote it out like this 0 = - …
  5. Trigonometry

    Solve for x: 1. 2sin(2x)+cos(x)=0 2. cos(2x)=-2sin(x) 3. tan(x)=2sin(x)
  6. trigonometry (please double check this)

    Solve the following trig equations. give all the positive values of the angle between 0 degrees and 360 degrees that will satisfy each. give any approximate value to the nearest minute only. 1. sin2ƒÆ = (sqrt 3)/2 2. sin^2ƒÆ = …
  7. URGENT - Trigonometry - Identities and Proofs

    Okay, today, I find myself utterly dumbfounded by these three questions - Write a proof for - 2/(sqrt(3)cos(x) + sin(x))= sec((pi/6)-x) Solve the following equation - 2sin(2x) - 2sin(x) + 2(sqrt(3)cos(x)) - sqrt(3) = 0 Find all solutions …
  8. Math

    Prove each identity: a) 1-cos^2x=tan^2xcos^2x b) cos^2x + 2sin^2x-1 = sin^2x I also tried a question on my own: tan^2x = (1 – cos^2x)/cos^2x R.S.= sin^2x/cos^2x I know that the Pythagorean for that is sin^2x + cos^2x That's all I …
  9. trigonometry

    Solve the equation in the interval [0, 2pi]. 2sin(t)cos(t)-cos(t)+2sin(t)-1=0 Answer needs to be formatted as x={Insert answers here}pi If the answer is something like pi/6 it simply goes in as 1/6th since pi is already given at the …
  10. another please help me check~calculus maths

    y=3e^(2x)cos(2x-3) verify that d^2y/dx^2-4dy/dx+8y=0 plz help me i tried all i could but it become too complicated for me here set u=3e^(2x) v=cos(2x-3) du/dx=6e^(2x) i used chain rule dv/dx=-2sin(2x-3) dy/dx=-3e^(2x)sin(2x-3)+cos(2x-3)6e^(2x) …

More Similar Questions