arcsin(x^2 −4) = arcsin(2x + 4) ⇒ x = ?


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
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
	$arcsin(x^2 −4) = arcsin(2x+4) ⇔ { ((−1 ≤x^2 −4 ≤ +1)),((−1 ≤ 2x+4 ≤ +1)),((x^2 −4 = 2x+4)) :} ⇔ { ((x∈[−(√5);−(√3)]∪[+(√3);+(√5)])),((x∈[−(5/2);−(3/2)])),((x^2 −2x−8 = 0)) :} x^2 −2x−8 = (x+2)(x−4) = 0 x = −2 or x = 4 but 4∉[−(5/2);−(3/2)] S = {−2}$
	$\arcsin (x^{2} - 4) = \arcsin (2 x + 4)$ $\Leftrightarrow {\begin{cases} - 1 ⩽ x^{2} - 4 ⩽ + 1 \\ - 1 ⩽ 2 x + 4 ⩽ + 1 \\ x^{2} - 4 = 2 x + 4 \end{cases}$ $\Leftrightarrow {\begin{cases} x \in [- \sqrt{5}; - \sqrt{3}] \cup [+ \sqrt{3}; + \sqrt{5}] \\ x \in [- \frac{5}{2}; - \frac{3}{2}] \\ x^{2} - 2 x - 8 = 0 \end{cases}$ $x^{2} - 2 x - 8 = (x + 2) (x - 4) = 0$ $x = - 2 or x = 4 but 4 \notin [- \frac{5}{2}; - \frac{3}{2}]$ $S = {- 2}$
	Commented by mathdanisur last updated on 16/Jul/21

	$t h a n k y o u S e r c o o l$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum