Could someone please explain to me why in 6sinx = 5 - 8 cosx, x can only equal 96°52' and 336°52'? Why can't it also be 23°8' and 263°8'???

Results 1 to 5 of 5

- September 10th 2008, 03:59 AM #1

- Joined
- Jul 2008
- Posts
- 347

- September 10th 2008, 04:47 AM #2

- Joined
- Apr 2005
- Posts
- 1,631

Because when you plug those last two angles in the original equation, the equation is false.

The first two make the original equation true.

-------------

6sinX = 5 -8cosX

Square both sides,

36sin^2(X) = 25 -80cosX +64cos^2(X)

36(1 -cos^2(X)) = 25 -80cosX +64cos^2(X)

100cos^2(X) -80cosX -11 = 0

Using the quadradtic formula,

cosX = 0.91962 or -0.11962

When cosX = 0.91962 .......positive cosine value, so X is in the 1st or 4th quadrants,

X = 23.1294 deg .....1st quadrant

X = 336.8706 deg ....4th quadrant

When cosX = -0.11962 ......negative cosine value, so X is in the 2nd or 3rd quadrants,

X = 96.8702 deg .....2nd quadrant

X = 263.1298 deg ....3rd quadrant

Since X = 23.1294 deg and X = 263.1298 deg do not make the original equation true, then it means X is not in the 1st and 3rd quadrants.

- September 11th 2008, 12:59 AM #3

- Joined
- Jul 2008
- Posts
- 347

- September 16th 2008, 01:58 AM #4

- Joined
- Jul 2008
- Posts
- 347

- September 16th 2008, 02:12 AM #5

- Joined
- Jun 2008
- Posts
- 792