∣ x+(√(1−x^2 )) ∣ = (√2) (2x^2 −1 ) find solution


Question and Answers Forum

All Questions Topic List
Arithmetic Questions
Previous in All Question Next in All Question
Previous in Arithmetic Next in Arithmetic
Question Number 134670 by benjo_mathlover last updated on 06/Mar/21

$∣ x + \sqrt{1 - x^{2}} ∣ = \sqrt{2} ({2 x}^{2} - 1)$ $find solution$
	Answered by EDWIN88 last updated on 06/Mar/21
	$(1) 1−x^2 ≥ 0 ⇒ −1≤x≤1 (2) let x = cos t ; t∈ [ 0,π ] ∣cos t +(√(1−cos^2 t)) ∣ =(√2) (2cos^2 t−1) ∣cos t +∣sin t∣∣ = (√2) cos 2t ∣ cos t+sin t ∣ = (√2) cos 2t (cos t+sin t)^2 = 2cos^2 2t ; { ((cos 2t ≥ 0)),((0≤t≤π)) :} (i) 1+sin 2t = 2−2sin^2 2t 2sin^2 2t +sin 2t−1 = 0→ { ((sin 2t=−1)),((sin 2t=(1/2))) :} { ((t=−(π/4)+kπ)),((t=(π/(12))+kπ ; t=((5π)/(12))+kπ)) :}→ t=(π/(12)) ; t=((3π)/4) { ((x = cos (π/(12))= (√((2+(√3))/4)) = ((√(2+(√3)))/2))),((x=cos ((3π)/4)=−((√2)/2))) :} therefore the solution is determinant (((x=((√(2+(√3)))/2))),((x =−((√2)/2))))$
	$(1) 1 - x^{2} ⩾ 0 \Rightarrow - 1 ⩽ x ⩽ 1$ $(2) let x = \cos t; t \in [0, π]$ $∣ \cos t + \sqrt{1 - \cos^{2} t} ∣ = \sqrt{2} (2 \cos^{2} t - 1)$ $∣ \cos t + ∣ \sin t ∣∣ = \sqrt{2} \cos 2 t$ $∣ \cos t + \sin t ∣ = \sqrt{2} \cos 2 t$ ${(\cos t + \sin t)}^{2} = 2 \cos^{2} 2 t; {\begin{cases} \cos 2 t ⩾ 0 \\ 0 ⩽ t ⩽ π \end{cases}$ $(i) 1 + \sin 2 t = 2 - 2 \sin^{2} 2 t$ $2 \sin^{2} 2 t + \sin 2 t - 1 = 0 \to {\begin{cases} \sin 2 t = - 1 \\ \sin 2 t = \frac{1}{2} \end{cases}$ ${\begin{cases} t = - \frac{π}{4} + k π \\ t = \frac{π}{12} + k π; t = \frac{5 π}{12} + k π \end{cases} \to t = \frac{π}{12}; t = \frac{3 π}{4}$ ${\begin{cases} x = \cos \frac{π}{12} = \sqrt{\frac{2 + \sqrt{3}}{4}} = \frac{\sqrt{2 + \sqrt{3}}}{2} \\ x = \cos \frac{3 π}{4} = - \frac{\sqrt{2}}{2} \end{cases}$ $therefore the solution is \begin{array}{cc} x = \frac{\sqrt{2 + \sqrt{3}}}{2} \\ x = - \frac{\sqrt{2}}{2} \end{array}$
	Commented by john_santu last updated on 06/Mar/21

	$i t h i n k i t s h o u l d b e t \in [0, 2 π]$ $t h e n t = \frac{7 π}{4} \land t = \frac{13 π}{12} i s t h e s o l u t i o n$
	Answered by benjo_mathlover last updated on 06/Mar/21
	$∣x+(√(1−x^2 )) ∣ = (√2) (2x^2 −1) ⇒x+(√(1−x^2 )) = ±(√2) (2x^2 −1) ⇒(√(1−x^2 )) = −x±(√2)(2x^2 −1) squaring ⇒1−x^2 = x^2 ±2(√2) x(2x^2 −1)+2(2x^2 −1)^2 ⇒0 = 2x^2 −1±2(√2)x(2x^2 −1)+2(2x^2 −1)^2 ⇒0=(2x^2 −1)[1±2x(√2) +2(2x^2 −1)] (i) 2x^2 =1 ; x_(1,2) =± ((√2)/2) → { ((≈0.707)),((≈−0.707)) :} (ii) 4x^2 ±2(√2) x−1= 0 → { ((x_(3,4) = ((−2(√2) ± 2(√6))/8)=((−(√2) ±(√6))/4) { ((≈0.25)),((≈−0.965)) :})),((x_(5,6) = (((√2) ±(√6))/4) → { ((≈0.965)),((≈−0.25 )) :})) :} (iii)for (√(1−x^2 )) defined on −1≤x≤1 (iv) 2x^2 −1≥0 ; x≤−((√2)/2) ∪ x≥((√2)/2) solution is {±((√2)/2); ((−(√2) ±(√6))/4) ; (((√2) ±(√6))/4) }$
	$∣ x + \sqrt{1 - x^{2}} ∣ = \sqrt{2} ({2 x}^{2} - 1)$ $\Rightarrow x + \sqrt{1 - x^{2}} = \pm \sqrt{2} ({2 x}^{2} - 1)$ $\Rightarrow \sqrt{1 - x^{2}} = - x \pm \sqrt{2} ({2 x}^{2} - 1)$ $squaring$ $\Rightarrow 1 - x^{2} = x^{2} \pm 2 \sqrt{2} x ({2 x}^{2} - 1) + 2 {({2 x}^{2} - 1)}^{2}$ $\Rightarrow 0 = {2 x}^{2} - 1 \pm 2 \sqrt{2} x ({2 x}^{2} - 1) + 2 {({2 x}^{2} - 1)}^{2}$ $\Rightarrow 0 = ({2 x}^{2} - 1) [1 \pm 2 x \sqrt{2} + 2 ({2 x}^{2} - 1)]$ $(i) {2 x}^{2} = 1; x_{1, 2} = \pm \frac{\sqrt{2}}{2} \to {\begin{cases} \approx 0.707 \\ \approx - 0.707 \end{cases}$ $(ii) {4 x}^{2} \pm 2 \sqrt{2} x - 1 = 0$ $\to {\begin{cases} x_{3, 4} = \frac{- 2 \sqrt{2} \pm 2 \sqrt{6}}{8} = \frac{- \sqrt{2} \pm \sqrt{6}}{4} {\begin{cases} \approx 0.25 \\ \approx - 0.965 \end{cases} \\ x_{5, 6} = \frac{\sqrt{2} \pm \sqrt{6}}{4} \to {\begin{cases} \approx 0.965 \\ \approx - 0.25 \end{cases} \end{cases}$ $(iii) for \sqrt{1 - x^{2}} defined on - 1 ⩽ x ⩽ 1$ $(iv) {2 x}^{2} - 1 ⩾ 0; x ⩽ - \frac{\sqrt{2}}{2} \cup x ⩾ \frac{\sqrt{2}}{2}$ $solution is {\pm \frac{\sqrt{2}}{2}; \frac{- \sqrt{2} \pm \sqrt{6}}{4}; \frac{\sqrt{2} \pm \sqrt{6}}{4}}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum