1-2x-1-x-2-2-2x-2-1-To-solve-in-R-

Question Number 64631 by Mikael last updated on 19/Jul/19

\sqrt{\frac{1 + 2 x \sqrt{1 - x^{2}}}{2}} + 2 x^{2} = 1

To solve in R

Commented by behi83417@gmail.com last updated on 20/Jul/19

- 1 < x ⩽ 1, let : x = cost

\sqrt{\frac{1 + 2 cost . ∣ sint ∣}{\sqrt{2}}} + {2 \cos}^{2} t = 1

1 . sint ⩾ 0 \Rightarrow \sqrt{\frac{1 + 2 sintcost}{2}} + {2 \cos}^{2} t = 1

\Rightarrow \frac{∣ sint + cost ∣}{\sqrt{2}} + {2 \cos}^{2} t - 1 = 0

∣ \cos (t - \frac{π}{4}) ∣= - \cos 2 t = \cos (π - 2 t)

\overset{t ⩾ \frac{π}{4}}{\Rightarrow} t - \frac{π}{4} = 2 n π \pm (π - 2 t)

\Rightarrow {\begin{cases} t = (2 n + 1) \frac{π}{3} + \frac{π}{12} \Rightarrow t = \frac{5 π}{12} \\ t = (- 2 n + 1) π - \frac{π}{4} \Rightarrow t = \frac{3 π}{4} \end{cases}

\Rightarrow x = cost = - \frac{\sqrt{2}}{2}, \frac{\sqrt{6} - \sqrt{2}}{4} .

2 . sint ⩽ 0 \Rightarrow \frac{∣ cost - sint ∣}{\sqrt{2}} = - \cos 2 t

\Rightarrow∣ \cos (t + \frac{π}{4}) ∣= \cos (π - 2 t)

\Rightarrow {\begin{cases} t + \frac{π}{4} = 2 m π \pm (π - 2 t) \\ t + \frac{π}{4} = 2 m π \pm (2 t - π) \end{cases}

\dots \dots \dots \dots t = \frac{π}{12}, \frac{π}{4} \Rightarrow x = cost = \frac{\sqrt{6} + \sqrt{2}}{4}, \frac{\sqrt{2}}{2} .

Commented by MJS last updated on 20/Jul/19

nice method but you have to verify your

solutions . not all of them fit the given equation

Commented by Mikael last updated on 22/Jul/19

G o d b l e s s y o u S i r .

Answered by MJS last updated on 19/Jul/19

\frac{\sqrt{1 + 2 x \sqrt{1 - x^{2}}}}{\sqrt{2}} + 2 x^{2} = 1

\sqrt{1 + 2 x \sqrt{1 - x^{2}}} = \sqrt{2} - 2 \sqrt{2} x^{2}

squaring (will create false solutions!)

1 + 2 x \sqrt{1 - x^{2}} = 8 x^{4} - 8 x^{2} + 2

2 x \sqrt{1 - x^{2}} = 8 x^{4} - 8 x + 1

squaring (will create false solutions!)

4 x^{2} - 4 x^{4} = 64 x^{8} - 128 x^{6} + 80 x^{4} - 16 x^{2} + 1

x^{8} - 2 x^{6} + \frac{21}{16} x^{4} - \frac{5}{16} x^{2} + \frac{1}{64} = 0

x = \pm \sqrt{y}

y^{4} - 2 y^{3} + \frac{21}{16} y^{2} - \frac{5}{16} y + \frac{1}{64} = 0

y = z + \frac{1}{2}

z^{4} - \frac{3}{16} z^{2} = 0

\Rightarrow z_{1} = - \frac{\sqrt{3}}{4} z_{2} = 0 z_{3} = \frac{\sqrt{3}}{4}

\Rightarrow y_{1} = \frac{1}{2} - \frac{\sqrt{3}}{4} y_{2} = \frac{1}{2} y_{3} = \frac{1}{2} + \frac{\sqrt{3}}{4}

\Rightarrow x_{1} = \frac{\sqrt{2} - \sqrt{6}}{4} x_{2} = - \frac{\sqrt{2}}{2} x_{3} = - \frac{\sqrt{2} + \sqrt{6}}{4}

x_{4} = \frac{\sqrt{6} - \sqrt{2}}{4} x_{5} = \frac{\sqrt{2}}{2} x_{6} = \frac{\sqrt{2} + \sqrt{6}}{4}

but only x_{2} and x_{4} solve the given equation

x = - \frac{\sqrt{2}}{2} \lor x = \frac{\sqrt{6} - \sqrt{2}}{4}

Commented by Mikael last updated on 22/Jul/19

N i c e s o l u t i o n S i r .

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