Solve the equation on the interval [0,2pi).

2sin^2x-3sinx+1=0
(2sinx+1)(sinx+1)
I don't think I did the factoring correctly. When I multiply it out to double check I get 2sin^2x+3sinx+1

2 answers

you just got the signs reversed

(2 sinx - 1)(sinx - 1)
Notice that the original had a negative term.
Your factored form is all positive, so you know for sure it is not correct.

how about
(2sinx - 1)(sinx - 1) = 0 ? , yes that works
so sinx = 1/2 or sinx = 1

let's do the easy one, sinx x = 1 , so x = π/2

for sinx = 1/2, we know that sine is positive in quads I and II by the CAST rule
so x = π/6 or x = π-π/6 = 5π/6

so x = π/2, π/6, 5π/6
Similar Questions
  1. y= 2sin^2 xy=1- sinx find values of x inthe interval 0<x<360 if 2sin^x = 1-sinx this can be arranged into the quadratic. 2sin^2
    1. answers icon 0 answers
  2. solve each equation for 0=/<x=/<2pisin^2x + 5sinx + 6 = 0? how do i factor this and solve? 2sin^2 + sinx = 0 (2sinx - 3)(sinx
    1. answers icon 7 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
  4. The equation 2sinx+sqrt(3)cotx=sinx is partially solved below.2sinx+sqrt(3)cotx=sinx sinx(2sinx+sqrt(3)cotx)=sinx(sinx)
    1. answers icon 7 answers
more similar questions