Question

2sin^2 theta - sin theta - 1 = 0

Answer 1

edit: Find all solutions in the interval [0, 2pi)

Answer 2

\[2 \sin ^{2}\theta-\sin \theta-1=0\]
if p = sin theta,
2p^2-p-1=0

and then find the factor.

Answer 3

What is the best way to find the factor?

Answer 4

check getting two numbers a and b so that a+b=-1 and a*b= -1
- so but if there is 2p2 -p -1 =0 divide by 2 and will get p2 -p/2 -1/2 =0 so now you need getting two numbers a and b like a+b=-1/2 and a*b = -1/2

Answer 5

but if you solve this quadratic equation 2p2 -p -1=0 so will get
p_1,_2=(1+/- sqrt(1+8))/4 =(1+/- 3)/4 = - 2/4 = -1/2 and 1 so

Answer 6

so factorized will be (p-1)(p+1/2)=(p-1)(2p+1)

Answer 7

- so now you know that p= sin ß and hence will be (sinß -1)(2sinß+1)=0
so sinß-1=0 so sinß=1and 2sinß+1=0 so 2sinß=-1 so sinß= -1/2

Answer 8

ok ?

Answer 9

Thanks for breaking it down extensively. Appreciate it