04-01-11 Real Solution

\hspace{-16}\mathbf{\mathbb{F}}$ind real Solution of system of equations\\\\ $\mathbf{\sin x+2\sin (x+y+z)=0}$\\\\ $\mathbf{\sin y+3\sin (x+y+z)=0}$\\\\ $\mathbf{\sin z+4\sin (x+y+z)=0}$

P.S = Sorry Admin for posting Here (bcz it is not shown in Algebra forum)

@shubhodip last equation is √4z-1 or √4y-1 ?

If last equation is √4y-1 . then i am getting

x=y=z=1/2

and any other solution exists

Thanks

Assuming the prob is as man111 wrote:

If x≥y, then

\sqrt{4z-1} = x+y \ge 2y \Rightarrow 4z \ge 1+4y^2 \ge 4y \Rightarrow z \ge y

However using the same method eqn 3 yields either

y \ge x \ \text{or} \ y \ge z

So either z≥x=y or x≥y=z.

Suppose x=y, then we have from 2,

(x+z)^2 = 4x-1 and since 4x^2-4x+1 \ge 0

we have 4x^2 \ge (x+z)^2 \Rightarrow x \ge x

and so x=y=z.

Similarly the other case.

So x=y=z=1/2

or
(x+y)^2 = 4z-1 \Rightarrow 4z = 1+(x+y)^2 \ge 2(x+y)\Rightarrow z \ge \frac{x+y}{2}

Adding up the three similar inequalities we get

x+y+z \ge x+y+z which means equality must occur in all three cases.

which means x+y = y+z = z+x =1 and so x=y=z=1/2

wonderful sir!!!!

Yes Ketan you are right i meant √4y-1.

Multiplying all the equations by 2 and adding them up we get,
(4x-2\sqrt{4x-1})+(4y-2\sqrt{4y-1})+(4z-2\sqrt{4z-1})=0
So this is equivalent to,
(\sqrt{4x-1}-1)^2+(\sqrt{4y-1}-1)^2+(\sqrt{4z-1}-1)^2=0

So we get easily from here x=1/2 y=1/2 and z=1/2