arcsin-x-2-4-arcsin-2x-4-x-

Question Number 146865 by mathdanisur last updated on 16/Jul/21

a r c s i n (x^{2} - 4) = a r c s i n (2 x + 4)

\Rightarrow x = ?

Answered by Olaf_Thorendsen last updated on 16/Jul/21

\arcsin (x^{2} - 4) = \arcsin (2 x + 4)

\Leftrightarrow {\begin{cases} - 1 ⩽ x^{2} - 4 ⩽ + 1 \\ - 1 ⩽ 2 x + 4 ⩽ + 1 \\ x^{2} - 4 = 2 x + 4 \end{cases}

\Leftrightarrow {\begin{cases} x \in [- \sqrt{5}; - \sqrt{3}] \cup [+ \sqrt{3}; + \sqrt{5}] \\ x \in [- \frac{5}{2}; - \frac{3}{2}] \\ x^{2} - 2 x - 8 = 0 \end{cases}

x^{2} - 2 x - 8 = (x + 2) (x - 4) = 0

x = - 2 or x = 4 but 4 \notin [- \frac{5}{2}; - \frac{3}{2}]

S = {- 2}

Commented by mathdanisur last updated on 16/Jul/21

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