Άρρητη εξίσωση 7

Πηγή: Amir Sofi Matematika (facebook)

$\sqrt {\frac{1}{2}\left( {1 - 4x\sqrt {1 - 4{x^2}} } \right)} = 1 - 8{x^2}$

$\sqrt {\frac{1}{2}\left( {1 - 4x\sqrt {1 - 4{x^2}} } \right)} = 1 - 8{x^2}\mathop \Leftrightarrow \limits^{x \in \left[ { - 1,1} \right]}$

$\left| {2x - \sqrt {1 - 4{x^2}} } \right| = \sqrt 2 \left( {1 - 8{x^2}} \right)\mathop \Leftrightarrow \limits^{2x = \cos y,y \in \left[ {0,\pi } \right]} \left| {\cos y - \sqrt {1 - {{\cos }^2}y} } \right| = \sqrt 2 \left( {1 - 2{{\cos }^2}y} \right) \Leftrightarrow$

$\left| {\cos y - \left| {\sin y} \right|} \right| = - \sqrt 2 \cos 2y\mathop \Leftrightarrow \limits^{y \in \left[ {\frac{\pi }{4},\frac{{3\pi }}{4}} \right]} \left| {\cos \frac{\pi }{4}\cos y - \sin \frac{\pi }{4}\sin y} \right| = \cos \left( {\pi - 2y} \right) \Leftrightarrow$

$\left| {\cos \left( {\frac{\pi }{4} + y} \right)} \right| = \cos \left( {\pi - 2y} \right)\mathop \Leftrightarrow \limits^{\left( {\frac{\pi }{4} + y} \right) \in \left[ {\left[ {\frac{\pi }{2},\pi } \right]} \right]} \cos \left( {\frac{\pi }{4} + y} \right) = \cos 2y \Leftrightarrow y = \frac{\pi }{4} \vee y = \frac{{7\pi }}{{12}} \Leftrightarrow$

$x = \frac{{\sqrt 2 }}{4} \vee x = \frac{{\cos \frac{{7\pi }}{{12}}}}{2} = - \frac{{\sqrt 3 - 1}}{{4\sqrt 2 }}$

Άρρητη εξίσωση 7

Άρρητη εξίσωση 7

Re: Άρρητη εξίσωση 7