Question

Solve on the interval [0,2π) :
2sin^2 x - 3sin x + 1 = 0

Answer 1

@amistre64 or anyone else? got any ideas?

Answer 2

A. x = 2π, x = π/3
B. x = π, x = 2π/3, x = 5π/5
C. x = π/2, x = π/6, x = 5π/6
D. x = 2π, x = π/4, x = 5π/4

Answer 3

hint: let z = sin(x)

Answer 4

if you do that, then you can replace every copy of sin(x) with z to get this equation

2z^2 - 3z + 1 = 0

Answer 5

you would now solve for z, then use those solutions in terms of z to get the solutions in terms of x

Answer 6

z = 1
z = 1/2

Answer 7

good

Answer 8

so remember we let z = sin(x), so

if z = 1, then sin(x) = 1 ----> x = ???

Answer 9

if z = 1/2, then sin(x) = 1/2

which means

x = ??? or x = ???

Answer 10

you should get 3 solutions total (on this interval)

Answer 11

C. x = π/2, x = π/6, x = 5π/6

Answer 12

very good on all 3

Answer 13

Can you help with another one then? @jim_thompson5910

Answer 14

sure

Answer 15

4 csc x + 6 = -2 on the interval [0,2π)

Answer 16

first isolate csc(x)

Answer 17

replace it with z if you want like you did last time

Answer 18

x ≠ 0, z = -2/3

Answer 19

z = -2/x*

Answer 20

idk how you got that, if z = csc(x), then z = -2/x doesn't make much sense

Answer 21

whoops! z = -2

Answer 22

good

Answer 23

z = csc(x) as well

Answer 24

so because of those two facts, we know that

csc(x) = -2

Answer 25

and because csc(x) is really 1/sin(x), we can say

1/sin(x) = -2

sin(x) = -1/2 ... take the reciprocal of both sides

Answer 26

A. 7π/6, 11π/6
B. π/6, 5π/6
C. π/3, 5π/3
D. 2π/3, 4π/3

Answer 27

if sin(x) = -1/2, then

x = ???? or x = ???

Answer 28

humm... A?

Answer 29

B!

Answer 30

A or B, which one

Answer 31

A!

Answer 32

hopefully you're not just randomly guessing

Answer 33

No B. looked like the first

Answer 34

i see, they're slightly different

Answer 35

I recommend you using a unit circle if you're not sure

Answer 36

B - final answer!

Answer 37

nope

Answer 38

if sin(x) = -1/2, then x = 7pi/6 or x = 11pi/6