Question

Find the exact solution of cos2x+sinx=0 in the interval [0,2pi)
cos2x+sinx = 0
1 - 2sin^2x + sinx = 0
(2sinx+1)(sinx-1) = 0

sinx = -1/2 in QIII,IV
sinx = 1

I expect you can take it from there.

Related Questions

Solve the equation of the interval (0, 2pi) cosx=sinx I squared both sides to get :cos²x=sin²x... If sinx = 1/4 Find the exact value of cos2x I used the formula for double angle identities : cos2... Please help!!!!!!!!!!! Find all solutions to the equation in the interval [0,2π). 8. cos2x=cosx 10... Hello! Can someone please check and see if I did this right? Thanks! :) Directions: Find the ex...