1no bro
15 Answers
3tanx = 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?????
1dividing 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 !
1write 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
1dude wats in a name...hw did u get sin2x as 1????
sin2x+cos2x+2sinxcosx=1/4
sin2x=1/4-1
sin2x=-3/4
1cosx+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
1@ 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!!!)
1my answer (#4) has both the values.
one more thing to be noted here is that, both the values of tanx are negative.
hence sinx + cosx = 1/2 only when x ε Q2
hope the entire discussion is clear now :)
1sin x + cos x =1/2
√2 sin(x+45°)=1/2
sin(x+45°)=1/2√2
therefore tan(x+45°)=1/√7
(1 + tanx)/(1- tanx)=1/√7
therefore tanx=(√7-1)/(√7+1)
please correct me if i am wrong