30Oh oh .. I see! I'll try it again then! Basic error.
21 Answers
262@prophet sir - i knew that but actually we get the same values of x from that too.
1thnks but its not coming as you have neglected cos3\Theta=-cos(\Theta/2)
can there be any other substitution or would solving sin4xcos6x=(sin10x -sin2x)/2 and then doing certain adjustments be of some help??
but I had tried it and it was not coming.
341You have missed out on solutions given by
\cos 3 \theta = -\cos \frac{\theta}{2}
262i think x= 2cosθ will be a better substitution.
or, 2(4cos3θ - 3 cos θ ) = √(2(1+cosθ )
simplifying, we have cos 3θ =cos(θ /2)
now 3θ =2n.pi ±θ /2
principal values of θ =0, 4.pi/5 , 4.pi/7
hence x = 2 , 2cos 4.pi/5 , 2 cos 4.pi/7
21more hint:
There always exists some θ such that x = 2sinθ, now we get a trigonometric equation, which is a bit simpler to solve
1its easy to solve ∫sin4xcos6x but how to do the above one?
1ur solution is wrng manoj u cnt get 1 irrational solution
irrational solutions come in conjugate pairs
1Soln --- x3-3x = √x+2
=> x(x2-3) = √x+2
=> x2(x2-3)2 = x+2
=> x[x(x2-3)2-1] = 2
now
2 = 2 x 1
.: x = 2 or 1
putting x = 2 will satisfy the equation so the ans is 2
BUT THERE MAY BE A COMPLEX NO. SOLUTION TOO OF THE EQUATION
are u sure its a 3 degree polynomial?
considr the eqn
x=\sqrt{x+2}
is it 1 degree???
1Yeah , you considered the given equation to be a polynomial , whereas it is not so . Only after squaring each side , you get a polynomial , that too with extra roots due to squaring .
30Okay I'm tried my hand at this and came at an unexpected result.
By observation, one root turned out to be 2.
Being a 3rd degree equation, the equations should have 3 roots.
Let the roots be α, β and γ.
α=2
Now, since coefficient of x2 = 0,
β+γ=-2
now,
αβ+βγ+αγ=-3
α(β+γ) + βγ=-3
βγ=1
Therefore the roots must be the multiplicative inverses of each other.
Solving,
β=γ=-1
But, wierd thing is -1 is not a solution. Ricky bhaiya can you tell where I went wrong?
1What , is everybody sleeping ? : )
Come on , not every problem that I post is of olympiad level : (
Though that might be the truth : ) : )
1Sorry , but there are only three roots , and you have managed to find out two of them .
1at last i got the all ans
1. 2
2. - (√5+1)/2
3. (√5-1)/2
4. 7*(sqrt(3)*%i/2-1/2)/(9*(7*%i/(2*3^(3/2))+7/54)^(1/3))+(7*%i/(2*3^(3/2))+7/54)^(1/3)*(-sqrt(3)*%i/2-1/2)-1/3
5. (7*%i/(2*3^(3/2))+7/54)^(1/3)*(sqrt(3)*%i/2-1/2)+7*(-sqrt(3)*%i/2-1/2)/(9*(7*%i/(2*3^(3/2))+7/54)^(1/3))-1/3
6. (7*%i/(2*3^(3/2))+7/54)^(1/3)+7/(9*(7*%i/(2*3^(3/2))+7/54)^(1/3))-1/3
Sorry i cudnt rite in the readable form cz its too long and im lazy
1ya u are rite but this method can find on and at least one solution but sometimes not the one
1i think the other solutions are √5-12 and √5+12 am i correct
there can be more
1Sorry , but I did not say " x " has to be an integer . The factorising of " 2 " is not correct .