The Student Room Group

Even moar trig

Same interval for x as last time
0°≤x≤360°

4cosx(cosx-1)=-5cosx

i did
4cos²x-4cosx=-5cosx
4cos²x+cosx=0
cosx(4cosx+1)=0

but there's no answer in the back of the book which says that 0°,180°or 360° are answers
Reply 1
Original post by thefatone
Same interval for x as last time
0°≤x≤360°

4cosx(cosx-1)=-5cosx

i did
4cos²x-4cosx=-5cosx
4cos²x+cosx=0
cosx(4cosx+1)=0

but there's no answer in the back of the book which says that 0°,180°or 360° are answers


That would be because cosx=0\cos x = 0 does not yield 0,1800^{\circ},180^{\circ} or 360360^{\circ} as solution, that would be sinx=0\sin x = 0. cosx=0\cos x = 0 when x=90x = 90^{\circ} and I'll let you find the other solution.
Reply 2
Original post by thefatone
Same interval for x as last time
0°≤x≤360°

4cosx(cosx-1)=-5cosx

i did
4cos²x-4cosx=-5cosx
4cos²x+cosx=0
cosx(4cosx+1)=0

but there's no answer in the back of the book which says that 0°,180°or 360° are answers


You have done it correctly but cosx is not 0 when x=0,180 or 360
Reply 3
Thanks a bunch all i'll probably be posting loads more threads since i'll probably need more help on these xD
Reply 4
Original post by thefatone
Thanks a bunch all i'll probably be posting loads more threads since i'll probably need more help on these xD


I am too late to help?
Reply 5
Original post by thefatone
Thanks a bunch all i'll probably be posting loads more threads since i'll probably need more help on these xD


De nada.

Original post by TeeEm
I am too late to help?


You snooze you lose :wink:
Reply 6
Original post by Zacken
De nada.



You snooze you lose :wink:


I am glad you are doing more than your fair bit as I feel very "cannot be bothered" recently:smile:
Reply 7
My girl bled a bit. She had the best time tho in missionary, doggy style was orite too
Reply 8
Original post by TeeEm
I am glad you are doing more than your fair bit as I feel very "cannot be bothered" recently:smile:


No problemo! You deserve a break. :smile:

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