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Question Number 193676 by lapache last updated on 18/Jun/23

Solve  arcsin(2x)+arcsin(x(√3))=arcsin(x)

$${Solve} \\ $$$${arcsin}\left(\mathrm{2}{x}\right)+{arcsin}\left({x}\sqrt{\mathrm{3}}\right)={arcsin}\left({x}\right) \\ $$

Answered by Frix last updated on 18/Jun/23

sin^(−1)  t ∈R ⇔ −1≤t≤1  f(x)=sin^(−1)  2x +sin^(−1)  (√3)x −sin^(−1)  x  −(1/2)≤x≤(1/2)  f′(x)=(2/( (√(1−4x^2 ))))+((√3)/( (√(1−3x^2 ))))−(1/( (√(1−x^2 ))))>0∀x∈(−(1/2); (1/2))  ⇒ only solution is x=0

$$\mathrm{sin}^{−\mathrm{1}} \:{t}\:\in\mathbb{R}\:\Leftrightarrow\:−\mathrm{1}\leqslant{t}\leqslant\mathrm{1} \\ $$$${f}\left({x}\right)=\mathrm{sin}^{−\mathrm{1}} \:\mathrm{2}{x}\:+\mathrm{sin}^{−\mathrm{1}} \:\sqrt{\mathrm{3}}{x}\:−\mathrm{sin}^{−\mathrm{1}} \:{x} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\leqslant{x}\leqslant\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${f}'\left({x}\right)=\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} }}+\frac{\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{1}−\mathrm{3}{x}^{\mathrm{2}} }}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}>\mathrm{0}\forall{x}\in\left(−\frac{\mathrm{1}}{\mathrm{2}};\:\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\:\mathrm{only}\:\mathrm{solution}\:\mathrm{is}\:{x}=\mathrm{0} \\ $$

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