36
rahul
·2010-12-31 11:59:05
2 sin x + 3 cos x = 10
=> 2sinx = 10 - 3cosx
=> (2sinx)2 = (10 - 3cosx)2
=> 13cos2x - 60cosx + 96 = 0
which gives no real value of cos x
well i don't know furthur
2. I think its false as when we square an equation then some other root
gets involved...!
21
Shubhodip
·2010-12-31 21:50:31
calculate the maximum value of 2sinx + 3cosx like that
√13(2/√13 sinx + 3/√13 cosx)
= √13 (cos∂ sinx + sin∂cosx) where cos∂= 2/√13
= √13 sin (x+∂)
max and min value of sin(x+∂) is ±1
so the given expression lies between ±√13
so no real root
11
Joydoot ghatak
·2010-12-31 22:42:03
1st one was too easy... since sinx and cosx can never give more than 1...
the sum can never b 10..
11
Joydoot ghatak
·2010-12-31 22:43:43
for the second one.. i agree with rahulmishra..
1
kunl
·2010-12-31 22:51:49
well answer for the second one is true.
can anyone explain it?
1
Anirudh Kumar
·2011-01-01 02:50:34
for the second one
suppose you were given sinx = 1/9
the √sinx = ±1/3
but in the question it is given sinx= 1/9 and √sinx= +1/3 whixh are same.
i feel that's why.
1
aruntm
·2011-01-01 04:26:22
√sinx=(sinx)12=13
squaring it we get
((sinx)12)2=(13)2
→ sinx = 19
i think its like this...anyone else feels the same way??its the same explanation anirudh gave....