Post a New Question

trig

posted by .

Find all solutions in the interval [0,2pi].

sin 2x + sin x = 0

  • trig -

    sin 2 x = 2 sin x cos x

    2 sin x cos x + sin x = 0

    sin x (2 cos x + 1) = 0
    so
    x = 0 is a solution and x = 180 is a solution
    then
    cos x = -1/2
    that is at x = 120 and x = 240

  • trig -

    We should also include 360° , (from sinx = 0)

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Calculus

    I need to find the exact solutions on the interval [0,2pi) for: 2sin^2(x/2) - 3sin(x/2) + 1 = 0 I would start: (2sin(x/2)-1)(sin(x/2)-1) = 0 sin(x/2)=1/2 and sin(x/2)=1 what's next?
  2. math

    find all solutions in the interval [0,2 pi) sin(x+(3.14/3) + sin(x- 3.14/3) =1 sin^4 x cos^2 x Since sin (a+b) = sina cosb + cosb sina and sin (a-b) = sina cosb - cosb sina, the first problem can be written 2 sin x cos (pi/3)= sin …
  3. trig

    I need to find all solutions of the given equations for the indicated interval. Round solutions to three decimal places if necessary. 1.) 3sin(x)+1=0, x within [0,2pi) 2.) 2sin(sq'd)(x)+cos(x)-1=0, x within R 3.) 4sin(sq'd)(x)-4sin(x)-1=0, …
  4. Pre-Calculus

    cos^2(x) + sin(x) = 1 - Find all solutions in the interval [0, 2pi) I got pi/2 but the answer says {0, pi/2, pi} Where does 0 and pi come from for solutions?
  5. Math Trig

    Find all solutions on the interval [0,2pi] for the following: 2sin^2(x)-5sin(x)=-3 cos^2(x)+sin(x)=1
  6. trig

    Find the exact solutions of the equation in the interval [0,2pi). sin(x/2)+cos(x)=0
  7. trig

    Solve cos x-1 = sin^2 x Find all solutions on the interval [0,2pi) a. x=pi, x=pi/2, x= 2pi/3 b. x=3pi/7, x=pi/2, x=2pi/3 c. x=3pi/7, x=3pi/2, x=3pi/2 d. x=pi, x=pi/2, x=3pi/2
  8. Math

    Can I please get some help on these questions: 1. How many solutions does the equation,2sin^2 2 θ = sin2θ have on the interval [0, 2pi]?
  9. Pre-Calc

    Find all solutions of the equation in the interval [0, 2pi). Show all work. sin(x+(pi/6))-sin(x-((pi/6))=1/2
  10. Trigonometry

    Solve the equation for solutions in the interval 0<=theta<2pi Problem 1. 3cot^2-4csc=1 My attempt: 3(cos^2/sin^2)-4/sin=1 3(cos^2/sin^2) - 4sin/sin^2 = 1 3cos^2 -4sin =sin^2 3cos^2-(1-cos^2) =4sin 4cos^2 -1 =4sin Cos^2 - sin=1/4 …

More Similar Questions

Post a New Question