Solve Sinx=Cos2x-1 for all values between 0 and 2pi

1 answer

cos 2x = 1 - 2sin^2 x
so, you have

2sin^2 x + sin x = 0
sin x (2sin x + 1) = 0
sinx = 0 --> x = 0,2pi
sinx = -1/2 --> x = 7pi/6,11pi/6
Similar Questions
  1. Hello! Can someone please check and see if I did this right? Thanks! :)Directions: Find the exact solutions of the equation in
    1. answers icon 1 answer
  2. Solve the equation of the interval (0, 2pi)cosx=sinx I squared both sides to get :cos²x=sin²x Then using tri indentites I came
    1. answers icon 4 answers
  3. the problem is2cos^2x + sinx-1=0 the directions are to "use an identity to solve each equation on the interval [0,2pi). This is
    1. answers icon 3 answers
    1. answers icon 5 answers
more similar questions