Trignometric identities question

[tempeste]

Hi,

the question is

2sin^2 + sinX -1 = 0

I think the next step is then

2(1-cos^2)+cos^2=0

Did I do something wrong because I cannot seem to continue the question?

 

[marlon]

Just replace sin(x) by a variable p.
the equations then is 2p² + p - 1 = 0

Solve this equation for p and then set these solutions equal to sin(x)
At best you will have two goniometric equations to solve

marlon

 

[dextercioby]

Hi,

the question is

2sin^2 + sinX -1 = 0

I think the next step is then

2(1-cos^2)+cos^2=0

Did I do something wrong because I cannot seem to continue the question?

Yes,i've colored in red what is wrong,namely
\sin x -1=\cos^{2} x

That is wrong...As an indentity.It is valid as an equation in "x",which,i'm afraid,is not really equivalent to your initial one...

Daniel.

 

[tempeste]

I still don't really understand...

I tried it marlon's way

2p^2+p-1=0
2p^2+p=1
2p+p=1 (squareroot)
3p=1
p=1/3

Is the right?

 

[marlon]

I still don't really understand...

I tried it marlon's way

2p^2+p-1=0
2p^2+p=1
2p+p=1 (squareroot)
3p=1
p=1/3

Is the right?

Not at all.

Can you solve an equation like ax² + bx + c = 0 ?
Discriminant D = b² -4ac

first solution = \frac{-b + \sqrt{D}}{2a}
second solution = \frac{-b - \sqrt{D}}{2a}

Just apply this to your equation where x is now p

marlon

 

[HallsofIvy]

No, it isn't.

square root of 2p^2+ p is NOT 2p+ p!

Can you factor 2p^2+ p- 1?