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Question Number 158354 by cortano last updated on 03/Nov/21

\sqrt{\sin (\frac{π}{4} + x)} + \sqrt{\sin (\frac{π}{4} - x)} = \sqrt[4]{2 \cos 2 x}

Commented by tounghoungko last updated on 03/Nov/21

\sin (\frac{π}{4} + x) + \sin (\frac{π}{4} - x) + 2 \sqrt{\sin (\frac{π}{4} + x) \sin (\frac{π}{4} - x)} = \sqrt{2 \cos 2 x}

(i) \sin (\frac{π}{4} + x) + \sin (\frac{π}{4} - x) = 2 \sin \frac{π}{4} \cos x = \sqrt{2} \cos x

(i i) \sin (\frac{π}{4} + x) \sin (\frac{π}{4} - x) = - \frac{1}{2} (\cos \frac{π}{2} - \cos 2 x) = \frac{\cos 2 x}{2}

\Rightarrow \sqrt{2} \cos x + 2 \sqrt{\frac{\cos 2 x}{2}} = \sqrt{2 \cos 2 x}

\Rightarrow \sqrt{2} \cos x + \sqrt{2 \cos 2 x} = \sqrt{2 \cos 2 x}

\Rightarrow \cos x = 0 \Rightarrow x = \frac{π}{2}

c h e c k R H S \Rightarrow \sqrt[4]{2 \cos 2 x} = \sqrt[4]{2 \cos π} = \sqrt[4]{- 2} \notin R

s o t h e e q u a t i o n h a s n o s o l u t i o n

Commented by MJS_new last updated on 03/Nov/21

with x = n π + \frac{π}{2} lhs = rhs = \frac{1}{\sqrt[4]{2}} (1 + i) \Rightarrow you

found the solution!

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