conceptual doubt

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...!

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

1st one was too easy... since sinx and cosx can never give more than 1...
the sum can never b 10..

for the second one.. i agree with rahulmishra..

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.

√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....