Thread: Solve the Equation exactly over the interval [0, 2pi].

Solve the Equation exactly over the interval [0, 2pi]

the problems i do not understand are:
2sin^2x + sin x - 1 = 0
2sinx cosx + cos(x) = 0
2sin^2x - sin x = 0

if anyone could help me i would appreciate it alot. thanks

Originally Posted by trigonometry424

Solve the Equation exactly over the interval [0, 2pi]

the problems i do not understand are:
2sin^2x + sin x - 1 = 0
2sinx cosx + cos(x) = 0
2sin^2x - sin x = 0
2sin^2x + sin x - 1 = 0

if anyone could help me i would appreciate it alot. thanks

What have you tried?

1) Use the quadratic formula to get values of sin(x) and thus x.
2) Factor out cos(x)
3) Factor out sin(x)
4) Same as 1

If you're stuck post what you've tried

i know how to do problem 1 and 2 now i think i just need to know how to factor the 3rd one now

Originally Posted by trigonometry424

i know how to do problem 1 and 2 now i think i just need to know how to factor the 3rd one now

Unless I'm missing something equation 4 is the same as equation 1 so you've solved that too.

Number 3:


You can factor out sin(x) since it's a common factor:

Either or

Can you find x in those cases?

yea sorry it was an accident i edited my op

and would x = 1/2?

Originally Posted by trigonometry424

yea sorry it was an accident i edited my op

and would x = 1/2?

No. but not x.

It's probably easier to start with when and then - delete as appropriate depending on your boundary conditions

For we have as the principal solution but there is another one to find

ohh yea
so the answer would be pi/6 and 5pi/6 right?