Halp meh pliss, The number of positive values of x less than 360 degrees which satisfy the equation sin 1/2x =cos x is? a)0 b)1) c)2 d)3 e)4
You do math too? D:
Mujhe bhi yaqeen nahi horaha :P
looks like there are two http://www.wolframalpha.com/input/?i=sin%28x%2F2%29%2Ccos%28x%29+domain+0..2pi
write cosx in terms of sin^2 x/2 and solve!
1/2x=sin^2x/2 Right ?
cosx= 1-2sin^2 (x/2)
you get a quadratic.. solve for x..
2sin^2(x/2)+cosx-1=0
2t^2 + t -1=0 where t=sin(x/2)
Quad,breaking the middle term or completing the square method?
breaking the middle term.. you would get (2t-1)(t+1)=0 where t=sin(x/2)
eh,2t-1=0=)t=1/2 t+1=0=)t=-1
yup!
Sin(x/2)=1/2 Sin(x/2)=-1
so there would be 3 values of x/2 in 0 to 360.. 30,120,270. so x=60,240,540.. since you want from 0 to 360.. there would be 2 solns..
it was Sin x/2
if sin(x/2) = 1/2 , what is x? 0>x>360
Oh so, sin x/2 =cos x ?
1- sin^2(x/2)= cosx @hba
em BOREEEEEED.
look.. let x/2=k .. sin(k)=1/2 --> k=30, 120.. --> x=60, 240
cosx= 1-2sin^2 (x/2) @hba missed the two..
thik hai? @uri
@yrelhan4 since \(\displaystyle 0 \leq x \leq 360\) then \(\displaystyle 0 \leq \frac{x}{2} \leq 180\)
now, \(\sin x/2 = 1/2\) therefore \(x/2=30, 150\) or \(x=60, 300\)
Yrelhan GUESSSS WHAT?! I donyt undastand aaaanythang nd ima sho confuzzeld tooh,I caant eveen speel corractly. o.O
lol.. i am such a fool.. you are right! @sirm3d
Wut iz the aaansweeer?! none of my option matches your answer!
i just wrotw150 value wrong, right? why am i a fool then? :P
the other equation, \(\sin x/2 = -1\) has no solution in \(0 \leq \frac{x}{2} \leq 180\)
@uri 2.. final!
@sirm3d sinx/2=1 has no solution so we will leave it? xD
@uri its sin(x/2) = -1
k.
@uri, just count the number of solutions: \(x=60^0,\;300^0\)
@sirm3d 2 solutions?
yup. (C) 2
are kitni baar puchegi? @uri :P
AMG THANKYOUUUUUUUUUUUUU!AMG THANKYOUU SOO MUCHHHHH!!!! :D
mera medal.. :'( .. aadha to maine solve kiya..
huh!
You have 2 now :P
haha.. he's got 2 now too!
Happy? :D
ha! yes.. :P
Yw xD
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