The number of solutions of ∣x+3∣+∣x−2∣+3sinx−3=0 is


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Question Number 191149 by SLVR last updated on 19/Apr/23

$T h e n u m b e r o f s o l u t i o n s o f$ $∣ x + 3 ∣ + ∣ x - 2 ∣ + 3 s i n x - 3 = 0 i s$
Commented by SLVR last updated on 19/Apr/23

$s i r k i n d l y h e l p m e$
Commented by SLVR last updated on 19/Apr/23

$s i r M r . W . . . k i n d l y h e l p i f$ $y o u a r e f r e e$
Commented by Frix last updated on 19/Apr/23
I'm Freex ��
	Answered by Frix last updated on 19/Apr/23
	$∣x+3∣+∣x−2∣=3(1−sin x) f_1 (x)=∣x+3∣+∣x−2∣= { ((−2x−1; x<−3)),((5; −3≤x≤2)),((2x+1; x>2)) :} 0≤(f_2 (x)=3(1−sin x))≤6 x∈[−(7/2); (5/2)]: 5≤f_1 (x)≤6 ⇒ Solutions possible within this interval ⇒ 2 solutions: x_1 =−π+sin^(−1) (2/3) x_2 =−sin^(−1) (2/3)$
	$∣ x + 3 ∣ + ∣ x - 2 ∣= 3 (1 - \sin x)$ $f_{1} (x) =∣ x + 3 ∣ + ∣ x - 2 ∣= {\begin{cases} - 2 x - 1; x < - 3 \\ 5; - 3 ⩽ x ⩽ 2 \\ 2 x + 1; x > 2 \end{cases}$ $0 ⩽ (f_{2} (x) = 3 (1 - \sin x)) ⩽ 6$ $x \in [- \frac{7}{2}; \frac{5}{2}] : 5 ⩽ f_{1} (x) ⩽ 6 \Rightarrow$ $Solutions possible within this interval \Rightarrow$ $2 solutions :$ $x_{1} = - π + \sin^{- 1} \frac{2}{3}$ $x_{2} = - \sin^{- 1} \frac{2}{3}$
	Commented by SLVR last updated on 19/Apr/23

	$t h a n k s s i r$
	Answered by mr W last updated on 19/Apr/23

	$∣ x + 3 ∣ + ∣ x - 2 ∣= 3 (1 - \sin x)$ $f (x) = g (x)$ $L H S f (x) ⩾ 5$ $0 ⩽ R H S g (x) ⩽ 6$ $t h a t m e a n s : s o l u t i o n c o u l d e x i s t .$ $w i t h x < - 3 :$ $f (x) = - 3 - x + 2 - x = - 1 - 2 x$ $f o r x < - 3.5 f (x) > 6, t h a t m e a n s$ $f o r x < - 3.5 f (x) \neq g (x)$ $f o r - 3.5 ⩽ x < - 3 g (x) < 3 (1 + \sin 3) \approx 3.423 < 5$ $t h a t m e a n s f o r - 3.5 ⩽ x < - 3 f (x) \neq g (x)$ $\Rightarrow f o r x < - 3 f (x) = g (x) h a s n o s o l u t i o n .$ $w i t h x > 2 :$ $s i m i l a r l y t o a b o v e f (x) = g (x) h a s$ $n o s o l u t i o n .$ $w i t h - 3 ⩽ x ⩽ 2 :$ $f (x) = 5$ $f (x) = g (x)$ $5 = 3 (1 - \sin x)$ $\Rightarrow \sin x = - \frac{2}{3}$ $\Rightarrow x = - \sin^{- 1} \frac{2}{3} o r - π + \sin^{- 1} \frac{2}{3}$ $t o t a l l y :$ $∣ x + 3 ∣ + ∣ x - 2 ∣ + 3 s i n x - 3 = 0 h a s$ $2 s o l u t i o n s :$ $x = - \sin^{- 1} \frac{2}{3} o r - π + \sin^{- 1} \frac{2}{3}$
	Commented by SLVR last updated on 19/Apr/23

	$T h a n k s a l o t . . P r o f . W$ $g o d b l e s s y o u s i r$
	Answered by mehdee42 last updated on 19/Apr/23

	$l e t f (x) = ∣ x - 2 ∣ + ∣ x + 3 ∣ & g (x) = 3 - 3 s i n x$ $\forall x - 3 ⩽ x ⩽ 2 \to f (x) = 5 \Rightarrow 5 + 3 s i n x - 3 = 0 \Rightarrow s i n x = - \frac{2}{3} \Rightarrow x = - {s i n}^{- 1} (\frac{2}{3}) & - π + {s i n}^{- 1} (\frac{2}{3}) ✓$ $\forall x < - 3, x > 2 \Rightarrow f (x) > g (x) \Rightarrow t h e r e f o r t h e e q u a t i o n f (x) = g (x) h a s n o s o l u t i o n .$
	Commented by mehdee42 last updated on 19/Apr/23

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