If A, B, C are the angles of triangle show that system of equations
$x sin 2 A + y sin C + z sin B = 0$
$x sin C + y sin 2 B + z sin A = 0$
$x sin B + y sin A + z sin 2 C = 0$
possesses non-trivial solution.

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Solution

Write sin $2 A = 2 sin A cos A = 2 k a \frac{b^{2} + c^{2} - a^{2}}{2 b c}$

Now mutliply $R_{1}, R_{2}$ and $R_{3}$ by bc, ca, ab respectively and hence divide $△$ by $a^{2} b^{2} c^{2}$ .

Then take out a, b, c common from $C_{1}, C_{2} a n d C_{3}$ respectively.
$△ = \frac{k^{3}}{a b c} (a^{2} - b^{2}) (a^{2} - c^{2}) \times ∣ ∣ ∣ ∣ ∣ \begin{matrix} b^{2} + c^{2} - a^{2} & 1 & 1 c^{2} & 1 & 1 b^{2} & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0$ .
Since $△ = 0$ , the system has non-trivial solution.

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