trig

posted by .

Find all solutions of the equation 2sin^2x-cosx=1 in the interval [0,2pi)

x1= ?
x2=?
x3=?

  • trig -

    2 (1 - cos^2 x) - cos x = 1
    2 - 2 cos^2 x -cos x = 1
    2 cos^2 x + cos x -1 = 0
    let y = cos x
    then
    2 y^2 +y -1 = 0
    (y+1)(2y-1) = 0
    y = .5 or y = -1
    so
    cos x = .5 or cos x = -1
    so
    x = 60 degrees or 300 degrees
    or 180 degrees

  • trig -

    Rewrite as
    2(1-cos^2x)-cosx = 1.
    Rearrange as
    2cos^2x + cos x -1 = 0,
    which can be factored to provide:
    (2cosx -1)(cosx +1) = 0.
    This means
    cosx = 1/2 or -1
    Use that to find the three angles.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. 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, …
  2. 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
  3. trig

    Find ALL solutions to the equation on the interval [0, 2pie]: 2sin(1/2X) = 1
  4. Trigonometry

    Find all solutions of the equation in the interval [0,2pi] algebraically. sin^2x + cosx + 1 = 0
  5. calculus

    Find all solutions to the equation in the interval [0,2pi) Cosx-cos2x
  6. 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]?
  7. Math

    Directions: Find all solutions of the equation in the interval (0, 2pi) sin x/2=1-cosx
  8. Trig

    Find all solutions of the equation in the interval [0,2pi). 2sin theta+1=0. Write answer in radians in terms of pi.
  9. math

    Find all solutions to the equation tan(t)=1/tan (t) in the interval 0<t<2pi. Solve the equation in the interval [0, 2pi]. The answer must be a multiple of pi 2sin(t)cos(t) + sin(t) -2cos(t)-1=0 Find all solutions of the equation …
  10. Trig

    Find all solutions of the equation in the interval [0,2pi) 2 cos^2 x-cos x = 0 -2cos^2 + cosx + 0 (x+1/2) (x+0/2) (2x+1) (x+0) -1/2,0 2Pi/3, 4pi/3, pi/2, 3pi/2 my teacher circled pi/2 and 3pi/2 What did I do wrong?

More Similar Questions