Pre-Cal(Please check)


1) Find all the solutions of the equation in the interval (0,2pi).

sin 2x = -sqrt3/2

I know that the angles are 4pi/3 and 5pi/3 and then since it is sin I add 2npi to them.

2x = 4pi/6 + 2npi 2x = 5pi/3 + 2npi

Now I know that I have to divide by 2 but that is where I an not sure.

x = 4pi/6 + npi x=10pi/3 + npi

Is that correct???

asked by Hannah
  1. Here is how I do these:
    sin 2x = -√3/2
    so by the CAST rule, 2x must be in quadrants III or IV
    The angle in standard position (ignore the negative) is pi/3
    so for III, 2x = pi + pi/3 = 4pi/3
    for IV, 2x = 2pi - pi/3 + 5pi/3, you had both of those correct
    Then x = 2pi/3 and 5pi/6

    the period of sin 2x is pi, so adding/subtracting multiples pi to our two angles will yield more angles
    since your domain is (0,2pi)
    other solutions for x are 2pi/3 + pi = 5pi/3
    and 5pi/6 + pi = 11pi/6

    so x = 2pi/3, 5pi/3, 5pi/6 and 11pi/6
    we would have

    posted by Reiny

Respond to this Question

First Name

Your Response

Similar Questions

  1. Pre-Cal(Please check)

    1) Find all the solutions of the equation in the interval (0,2pi). sin 2x = -sqrt3/2 I know that the angles are 4pi/3 and 5pi/3 and then since it is sin I add 2npi to them. 2x = 4pi/6 + 2npi 2x = 5pi/3 + 2npi Now I know that I
  2. Pre-Cal (Please help)

    Find all the soultions of the equation in the interval (0,2pi) sin 2x = -sqrt3 /2 sin -sqrt3 /2 is 4pi/3 and 5pi/3 in the unit circle 2x= 4pi/3 + 2npi 2x=5pi/3 + 2npi I do not know what to do at this point
  3. Pre-Cal

    Find all the soultions of the equation in the interval (0,2pi) sin 2x = -sqrt3 /2 sin -sqrt3 /2 is 4pi/3 and 5pi/3 in the unit circle 2x= 4pi/3 + 2npi 2x=5pi/3 + 2npi I do not know what to do at this point.
  4. Math

    1. On the interval [0, 2pi] what are the solutions to the equation sin3xcos2x = -cos3xsin2x + 1? pi/10 and pi/2? 2. What is the value of tan75degrees? √(3) + 1)/(1 - √(3))? 3. Value of cos(130degrees)cos(130degrees) +
  5. Pre-Cal(Please check)

    Approximate the equation's solutions in the interval (0,2pi) sin2x sinx = cosx 2cos(x) (1/2-sin^2x) = 0 Then I got 3pi/2, pi/2, pi/6 and 5pi/6 Then I substituted 0-3 and got 3pi/2 , 5pi/2 , 9pi/2 , pi/2, pi/6, 7pi/6, 13pi/6 ,
  6. Pre-Cal

    Find all the solutions of the equation in the interval (0,2pip) sin(x + pi/6) - sin(x -pi/6) = 1/2 I am only stuck on the last part. I have 2sinx(sqrt3/2) = 1/2 Does the 2 cancel out with the sqrt 3/2. I am not sure what to do.
  7. Pre-Cal(Urgent!!)

    Find all the solutions of the equation in the interval (0,2pip) sin(x + pi/6) - sin(x -pi/6) = 1/2 I am only stuck on the last part. I have 2sinx(sqrt3/2) = 1/2 Does the 2 cancel out with the sqrt 3/2. I am not sure what to do.
  8. Math

    Find the solution. 1. sqrt2 sin x + 1 = 0 My ans: x = 5pi/4 + 2npi and 7pi/4 + 2npi 2. sec x - 2 = 0 My ans: x = pi/3 + 2npi and 5pi/3 + 2npi
  9. Precalculus check answers help!

    1. Which expression is equivalent to cos 2theta for all values of theta ? cos^2 theta – sin^2 theta ~ cos^2 theta – 1 1 – 2 sin^2 theta 2 sin theta cos theta 2. Use a half–angle identity to find the exact value of sin
  10. Pre-Cal(Please help)

    Approximate the equation's soultions in the interval (o, 2pi). If possible find the exact solutions. sin 2x sinx = cosx I do not know where to start.

More Similar Questions