x-1-x-2-2-2x-2-1-find-solution-

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 \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 + \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}}