A little help with Trigonometry please

[Ricochet J]

Hello. I have my Core2 Maths exam on Monday so i'm just going over the things I am weak on. And this is Trigonometry.
Right now, i'm pretty sure i've nearly nailed Trig, but one or two things are bugging me.

Q) Find all values of X in the interval 0 ≤ x < 360° for which
2sin²x - 2cosx - cos²x = 1

Now the mark scheme to that question shows this:
2(1 − cos2 x) − 2 cos x − cos2 x = 1
3 cos2 x + 2 cos x − 1 = 0
(3 cos x − 1)(cos x + 1) = 0
cos x = −1 or 1
3 A1
x = 180 or 70.5, 360 − 70.5 B2
x = 70.5° (1dp), 180°, 289.5° (1dp)



Now the thing I don't get is the first step when they factorise that equation, where did the sin²x dissapear to and why did they replace it with cos?

I tried to apply these identities to it:
tanθ = sinθ / cosθ and sin²θ + cos²θ = 1

But I still didn't get it!

Any help would be vastly appreciated. Thanks

 

[Galaxy]

sin²θ + cos²θ = 1

sin²θ = 1 - cos²θ

That should help you

 

[PhilthyPhil]

sin²x=1/2(1-cos2x)
cos²x=1/2(1+cos2x)

 

[Ricochet J]

OMG I can't believe I didn't even think about rearranging the identities!
Thank you Galaxy and PhilthyPhil you've made my day, honestly you have

I understand it now! Only when I do it I would say let cosx = y to neaten things up a bit, then when I have factorised and have y, I would say cosx = y
And go from there.

Thanks again

 

[spirit]

Hello. I have my Core2 Maths exam on Monday so i'm just going over the things I am weak on. And this is Trigonometry.
Right now, i'm pretty sure i've nearly nailed Trig, but one or two things are bugging me.

Q) Find all values of X in the interval 0 ≤ x < 360° for which
2sin²x - 2cosx - cos²x = 1

Now the mark scheme to that question shows this:
2(1 − cos2 x) − 2 cos x − cos2 x = 1
3 cos2 x + 2 cos x − 1 = 0
(3 cos x − 1)(cos x + 1) = 0
cos x = −1 or 1
3 A1
x = 180 or 70.5, 360 − 70.5 B2
x = 70.5° (1dp), 180°, 289.5° (1dp)



Now the thing I don't get is the first step when they factorise that equation, where did the sin²x dissapear to and why did they replace it with cos?

I tried to apply these identities to it:
tanθ = sinθ / cosθ and sin²θ + cos²θ = 1

But I still didn't get it!

Any help would be vastly appreciated. Thanks

Are you sure this is correctly copied?

2sin²x - 2cosx - cos²x = 1
2(1-cos²x) - 2cosx - cos²x = 1
3cos²x + 2cosx - 1 = 0
(3cosx - 1)(cosx + 1)

thus cosx = -1 or cosx = 1/3
x = 180 or x = 70.5 (or 360-theta)

 

[spirit]

sin²x=1/2(1-cos2x)
cos²x=1/2(1+cos2x)

oh and if you cant be bothered to learn these just rmember how they come about

c²x + s²x = 1
c²x - s²x = c2x

(and if you notice in your second line, its equiv to 4s² so i recon you copied it wrong)