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# P2 Trig watch

1. Hi, I need help with this question if you can please

Find the values of x in (0,360 degrees) for which 7cosx+sinx=5 giving each answer to 1 decimal place.

I don't understand the (0,360 degrees) bit of this question
2. (Original post by TheQueen1986)
Hi, I need help with this question if you can please

Find the values of x in (0,360 degrees) for which 7cosx+sinx=5 giving each answer to 1 decimal place.

I don't understand the (0,360 degrees) bit of this question
There are infinitely many x that solve your equation. They are asking only for those solutions that are in the range

0 < x < 360 degrees.
3. Oh I see! Thanks
4. Damn, I need help once again on this one.

I did as follows and I don't end up with the right answers. I know this because when I substitute the answers back into the original equation, they don't equal to 5.

7cosx+sinx=5
49cos(squared)x+sin(squared)x=25
49(1-sin(squared)x)+sin(squared)x=25
49-49sin(squared)x+sin(squared)x=25
-48sin(squared)x=-24
sin(squared)x=0.5
sinx=+ or - square root of 0.5
x=45 degrees, 315 degrees.
5. [QUOTE=TheQueen1986]Damn, I need help once again on this one.

I did as follows and I don't end up with the right answers. I know this because when I substitute the answers back into the original equation, they don't equal to 5.

7cosx+sinx=5
49cos(squared)x+sin(squared)x=25
QUOTE]
Your error occurs in the second line.
if you square both sides you will have a 14cosxsinx term.
compare 7cosx+sinx to Rcos(x-A) (*)
so 7cosx+sinx=RcosxcosA+RsinxsinA
RcosA=7 (1)
Rsin A=1 (2)

dividing (2) by (1) gives sinA/cosA=1/7 so Tan A=1/7
squaring (1) and (2) and adding gives
R^2(cos^2A+sin^2A)=49+1
so R^2=50
then to finish off use these values for R and A in (*)
to get
Rcos(x-A)=5
6. oh yeah, I forgot, this is like the common type they ask in exams isn't it? Thanks for your help!

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