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C3 Trig Question: Where am I going wrong?

Question and my working out has attached :smile:

(edited 9 years ago)
It should be 17cos(x + 1.08) = 17
Original post by morgan8002
It should be 17cos(x + 1.08) = 17


Oh okay

So then I would so

x + 1.08 = arccos(1)

x= 0, pi, 2pi...

And then what would I do?
Original post by creativebuzz
Oh okay

So then I would so

x + 1.08 = arccos(1)

x= 0, pi, 2pi...

And then what would I do?

x + 1.08 = 0, 2pi. (cos pi = -1)
Original post by morgan8002
x + 1.08 = 0, 2pi. (cos pi = -1)


Ah, I understand! thank you :smile:

Is it alright if you could see what else I need to do in this question? Because I managed to get 3 of the answers but apparently I'm still missing 2 other answers (1.40 and 4.88)

Original post by creativebuzz
Ah, I understand! thank you :smile:

Is it alright if you could see what else I need to do in this question? Because I managed to get 3 of the answers but apparently I'm still missing 2 other answers (1.40 and 4.88)



I would factorise rather than divide. You are at risk of losing solutions if you divide (like in this case).
6sinxcosx=sinx6\sin x\cos x = \sin x
(sinx)(6cosx1)=0(\sin x)(6\cos x - 1) = 0
Original post by morgan8002
I would factorise rather than divide. You are at risk of losing solutions if you divide (like in this case).
6sinxcosx=sinx6\sin x\cos x = \sin x
(sinx)(6cosx1)=0(\sin x)(6\cos x - 1) = 0


Ah okay, thanks! I assumed dividing was okay if it involved making tan, but I guess not! Thank you for your help :smile:

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