aieee

tanx = sinx / cosx //

sin 2 x + 1/4 - sin x = 1 - sin 2 x ...........

2sin 2 x - sin x - 3/4 = 0 ..........

(1 ± √7)/2 ....................................

sin x = 1-√7/2 ...................

but this means sin x is <0 n p in 0 - pi

and other val/ is >1......

where am i goin rong?????

dividing with cosx on both sides, we get

2tanx+2 = √(1+tan²x).............. (provided x≠Π/2)

=> 4tan²x+4+8tanx = 1+tan²x

3tan²x + 8tanx + 3 = 0

=> tanx = (-8 ± 2√7)/6 = (-4 ± √7)/3

correct me if i am wrong !

yup race...............even i got d same

write sin x = tan x/√1+tan²x & cos x = 1/√1+tan²x
(-4-√7)/3 + 1 = (√1+tan²x)/2
right hand side can't be -ve but solveing L.H.S it is -ve, so (-4 - √7)/3
can't be the answer. it was asked in aieee 2006 and the answer given was (-4 - √7)/3. how is it possible........
correct me if i have made a mistake

dude wats in a name...hw did u get sin2x as 1????

sin²x+cos²x+2sinxcosx=1/4
sin2x=1/4-1
sin2x=-3/4

wat is wrong in post #6

cosx+sinx=1/2
squaring both sides

1+ sin2x=1/4

sin2x=-3/4

therefore x >pi/2

2tanx/1+ tan²x =-3/4
2tanx=(-3/4)-(3/4)*tan²x
you get an equation

3tan²x+8tanx+3=0

tanx=-8±√64-36
6

therefore tanx<0

tanx=-8-√64-36
6
tanx=-4-√7
3

@ 8x10 regarding #6

secx = ±√1+tan²x ....... positive when x ε Q1 and negative when x ε Q2.

also tanx is positive in Q1 and negative in Q2. so while substituting sinx, we get positive for both quadrants Q1 and Q2; whereas, for cosx we get positive for Q1 and negative for Q2.

u have neglected this sign convention. (be cautious buddy!!!)

confirmation please.!