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Question Number 131595 by Algoritm last updated on 06/Feb/21

Commented by Algoritm last updated on 06/Feb/21

x_1 =37/2   x_2 =77/4   x_3 =65/4  Σx=37/2+77/4+65/4=54    prove that!

$$\mathrm{x}_{\mathrm{1}} =\mathrm{37}/\mathrm{2}\:\:\:\mathrm{x}_{\mathrm{2}} =\mathrm{77}/\mathrm{4}\:\:\:\mathrm{x}_{\mathrm{3}} =\mathrm{65}/\mathrm{4} \\ $$$$\Sigma\boldsymbol{\mathrm{x}}=\mathrm{37}/\mathrm{2}+\mathrm{77}/\mathrm{4}+\mathrm{65}/\mathrm{4}=\mathrm{54}\:\: \\ $$$$\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}}! \\ $$

Answered by mr W last updated on 06/Feb/21

sin^(−1) (sin ((πx)/2))=((π(17−x))/6)  −(π/2)≤((π(17−x))/6)≤(π/2)  ⇒14≤x≤20      ...(I)    sin ((πx)/2)=sin ((π(17−x))/6)  ⇒((πx)/2)=2kπ+((π(17−x))/6)  ⇒x=3k+((17)/4) with k=4, 5     ...(II)  i.e. x=16.25, 19.25    ⇒((πx)/2)=2kπ+π−((π(17−x))/6)  ⇒x=6k−((11)/2) with k=4    ...(III)  i.e. x=18.5    all solutions:  x=16.25, 18.5, 19.25  Σx=54

$$\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{sin}\:\frac{\pi{x}}{\mathrm{2}}\right)=\frac{\pi\left(\mathrm{17}−{x}\right)}{\mathrm{6}} \\ $$$$−\frac{\pi}{\mathrm{2}}\leqslant\frac{\pi\left(\mathrm{17}−{x}\right)}{\mathrm{6}}\leqslant\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{14}\leqslant{x}\leqslant\mathrm{20}\:\:\:\:\:\:...\left({I}\right) \\ $$$$ \\ $$$$\mathrm{sin}\:\frac{\pi{x}}{\mathrm{2}}=\mathrm{sin}\:\frac{\pi\left(\mathrm{17}−{x}\right)}{\mathrm{6}} \\ $$$$\Rightarrow\frac{\pi{x}}{\mathrm{2}}=\mathrm{2}{k}\pi+\frac{\pi\left(\mathrm{17}−{x}\right)}{\mathrm{6}} \\ $$$$\Rightarrow{x}=\mathrm{3}{k}+\frac{\mathrm{17}}{\mathrm{4}}\:{with}\:{k}=\mathrm{4},\:\mathrm{5}\:\:\:\:\:...\left({II}\right) \\ $$$${i}.{e}.\:{x}=\mathrm{16}.\mathrm{25},\:\mathrm{19}.\mathrm{25} \\ $$$$ \\ $$$$\Rightarrow\frac{\pi{x}}{\mathrm{2}}=\mathrm{2}{k}\pi+\pi−\frac{\pi\left(\mathrm{17}−{x}\right)}{\mathrm{6}} \\ $$$$\Rightarrow{x}=\mathrm{6}{k}−\frac{\mathrm{11}}{\mathrm{2}}\:{with}\:{k}=\mathrm{4}\:\:\:\:...\left({III}\right) \\ $$$${i}.{e}.\:{x}=\mathrm{18}.\mathrm{5} \\ $$$$ \\ $$$${all}\:{solutions}: \\ $$$${x}=\mathrm{16}.\mathrm{25},\:\mathrm{18}.\mathrm{5},\:\mathrm{19}.\mathrm{25} \\ $$$$\Sigma{x}=\mathrm{54} \\ $$

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