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Resoudre-dans-R-acos-x-bsin-x-c-x-0-sin-1-sin-x-d-1-d-1-

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Question Number 210231 by a.lgnaoui last updated on 03/Aug/24
Resoudre dans R { ((acos x−bsin x=c (x≠0))),((sin ((1/(sin x))) =d (−1≤d≤+1))) :}
$$\mathrm{Resoudre}\:\boldsymbol{\mathrm{dans}}\:\mathbb{R} \\ $$$$\begin{cases}{\boldsymbol{\mathrm{a}}\mathrm{cos}\:\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{b}}\mathrm{sin}\:\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{c}}\:\:\:\:\:\left(\boldsymbol{\mathrm{x}}\neq\mathrm{0}\right)}\\{\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{sin}\:\boldsymbol{\mathrm{x}}}\right)\:\:\:\:\:\:\:\:\:=\boldsymbol{\mathrm{d}}\:\:\:\:\left(−\mathrm{1}\leqslant\boldsymbol{\mathrm{d}}\leqslant+\mathrm{1}\right)}\end{cases} \\ $$$$ \\ $$
Commented by mr W last updated on 04/Aug/24
they are two different equations for variable x. they have different roots. they can not be an equation system! eqn. 1: a cos x−b sin x=c (√(a^2 +b^2 )) cos (x+tan^(−1) (b/a))=c assume ∣c∣≤(√(a^2 +b^2 )) cos (x+tan^(−1) (b/a))=(c/( (√(a^2 +b^2 )))) ⇒x+tan^(−1) (b/a)=2kπ±cos^(−1) (c/( (√(a^2 +b^2 )))) ⇒x=2kπ−tan^(−1) (b/a)±cos^(−1) (c/( (√(a^2 +b^2 )))) eqn. 2: sin ((1/(sin x)))=d ⇒(1/(sin x))=kπ+(−1)^k sin^(−1) d ⇒sin x=(1/(kπ+(−1)^k sin^(−1) d)) ⇒x=nπ+(−1)^n sin^(−1) ((1/(kπ+(−1)^k sin^(−1) d)))
$${they}\:{are}\:{two}\:{different}\:{equations}\:{for} \\ $$$${variable}\:{x}.\:{they}\:{have}\:{different}\:{roots}. \\ $$$${they}\:{can}\:{not}\:{be}\:{an}\:{equation}\:{system}! \\ $$$$ \\ $$$${eqn}.\:\mathrm{1}: \\ $$$${a}\:\mathrm{cos}\:{x}−{b}\:\mathrm{sin}\:{x}={c} \\ $$$$\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:\mathrm{cos}\:\left({x}+\mathrm{tan}^{−\mathrm{1}} \frac{{b}}{{a}}\right)={c} \\ $$$${assume}\:\mid{c}\mid\leqslant\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\mathrm{cos}\:\left({x}+\mathrm{tan}^{−\mathrm{1}} \frac{{b}}{{a}}\right)=\frac{{c}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$$\Rightarrow{x}+\mathrm{tan}^{−\mathrm{1}} \frac{{b}}{{a}}=\mathrm{2}{k}\pi\pm\mathrm{cos}^{−\mathrm{1}} \frac{{c}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$$\Rightarrow{x}=\mathrm{2}{k}\pi−\mathrm{tan}^{−\mathrm{1}} \frac{{b}}{{a}}\pm\mathrm{cos}^{−\mathrm{1}} \frac{{c}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$${eqn}.\:\mathrm{2}: \\ $$$$\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{sin}\:{x}}\right)={d} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{sin}\:{x}}={k}\pi+\left(−\mathrm{1}\right)^{{k}} \:\mathrm{sin}^{−\mathrm{1}} {d} \\ $$$$\Rightarrow\mathrm{sin}\:{x}=\frac{\mathrm{1}}{{k}\pi+\left(−\mathrm{1}\right)^{{k}} \:\mathrm{sin}^{−\mathrm{1}} {d}} \\ $$$$\Rightarrow{x}={n}\pi+\left(−\mathrm{1}\right)^{{n}} \mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{k}\pi+\left(−\mathrm{1}\right)^{{k}} \:\mathrm{sin}^{−\mathrm{1}} {d}}\right) \\ $$

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