1. ## Help?

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'???

2. Originally Posted by xwrathbringerx
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'???
Because when you plug those last two angles in the original equation, the equation is false.

The first two make the original equation true.

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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
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.

3. Is there any way of telling straight away which X would not hold the trig. equation true without subbing it into the trig. equation itself? (To save time during exams)