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Question Number 46277 by rahul 19 last updated on 23/Oct/18

Commented by rahul 19 last updated on 23/Oct/18

$t h i s i s p o s s i b l e o n l y w h e n s i n 3 x = 1$ $a n d s i n 7 x = 1 . . . . . .$ $. . . . .$
Commented by MJS last updated on 23/Oct/18

$yes! you already found the solution, all you$ $need is to calculate it$
	Answered by MJS last updated on 27/Oct/18

	$\sin 3 x = 1 \land - 2 π ⩽ x ⩽ 2 π \Rightarrow$ $\Rightarrow x = (- \frac{11}{6} + \frac{2 m}{3}) π \land m \in N \land 0 ⩽ m ⩽ 5$ $\sin 7 x = 1 \land - 2 π ⩽ x ⩽ 2 π \Rightarrow$ $\Rightarrow x = (- \frac{27}{14} + \frac{2 n}{7}) π \land n \in N \land 0 ⩽ n ⩽ 13$ $- \frac{11}{6} + \frac{2 m}{3} = - \frac{27}{14} + \frac{2 n}{7}$ $m = \frac{3 n - 1}{7} \Rightarrow n = 5 + 7 z \land m = 3 z + 2 \land z \in Z$ $0 ⩽ m ⩽ 5 \land 0 ⩽ n ⩽ 13 \Rightarrow z = 1 \land n = 13 \land m = 5$ $\Rightarrow one solution x = \frac{3 π}{2} \Rightarrow (A) is the right answer$
	Commented by rahul 19 last updated on 24/Oct/18

	$S i r, a n s . i s (B) i . e 2 s o l u t i o n s!$ $p l s s e x p l a i n l a s t 2 l i n e s . . . . . . .$ $m = \frac{3 n - 1}{7} . . . . w h a t i s z ? . . . . . h o w y o u$ $f o u n d r a n g e o f m, n ?$
	Commented by MJS last updated on 24/Oct/18

	$sorry I^{'} d been in a hurry . {there}^{'} s a 2^{nd} solution$ $with z = 0 \land n = 5 \land m = 2$
	Commented by MJS last updated on 24/Oct/18

	$range of m, n is clear because of the given$ $borders [- 2 π; 2 π]$ $m = \frac{3 n - 1}{7} \land m \in Z \Rightarrow \frac{3 n - 1}{7} \in Z \Rightarrow 7 ∣ (3 n - 1) \Rightarrow n \in {. . .; - 9; - 2; 5; 12; 19; . . .} \Rightarrow n = 5 + 7 z$
	Commented by rahul 19 last updated on 27/Oct/18

	$H e r e w e a r e u s i n g h i t a n d t r i a l m e t h o d,$ $r i g h t ? S o h o w w i l l y o u p r o v e n o o t h e r$ $s o l u t i o n e x i s t s e x c e p t t h e s e t w o . . . . ? ?$
	Commented by MJS last updated on 27/Oct/18

	$not hit and trial . we know - 2 ⩽ 2 \sin 3 x ⩽ 2 and$ $- 1 ⩽ \sin x ⩽ 1 . we need \sin 3 x = 1 \land \sin 7 x = 1$ $I generally solved both equations and looked$ $for which values of m and n the value of x is$ $in the demanded interval [- 2 π; 2 π] . then I$ $needed to find those values of x for which$ $both equations are true . I {can}^{'} t see any other$ $path . . .$
	Commented by MJS last updated on 27/Oct/18

	$you can compare the sets of solutions$ $\sin 3 x = 1 \Rightarrow$ $\Rightarrow x \in {- \frac{11 π}{6}; - \frac{7 π}{6}; - \frac{π}{2}; \frac{π}{6}; \frac{5 π}{6}; \frac{3 π}{2}}$ $\sin 7 x = 1 \Rightarrow$ $\Rightarrow x \in {- \frac{27 π}{14}; - \frac{23 π}{14}; - \frac{19 π}{14}; - \frac{15 π}{14}; - \frac{11 π}{14}; - \frac{π}{2}; - \frac{3 π}{14}; \frac{π}{14}; \frac{5 π}{14}; \frac{9 π}{14}; \frac{13 π}{14}; \frac{17 π}{14}; \frac{3 π}{2}; \frac{25 π}{14}}$
	Commented by rahul 19 last updated on 28/Oct/18
	thank you so much sir !��
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