sin-2-x-pi-4-sin-2-x-pi-4-7-cos-x-

Question Number 135497 by EDWIN88 last updated on 13/Mar/21

\sin^{2} (x + \frac{π}{4}) = \sin^{2} (x - \frac{π}{4}) + \sqrt{7} \cos x

Answered by john_santu last updated on 13/Mar/21

\Leftrightarrow \sin^{2} (x + \frac{π}{4}) - \sin^{2} (x - \frac{π}{4}) = \sqrt{7} \cos x

\Rightarrow \cos^{2} (\frac{π}{2} - x - \frac{π}{4}) - \sin^{2} (x - \frac{π}{4}) = \sqrt{7} \cos x

\Rightarrow \cos^{2} (x - \frac{π}{4}) - \sin^{2} (x - \frac{π}{4}) = \sqrt{7} \cos x

\Rightarrow \cos (2 x - \frac{π}{2}) = \sqrt{7} \cos x

\Rightarrow \sin 2 x = \sqrt{7} \cos x

\Rightarrow \cos x (2 \sin x - \sqrt{7}) = 0

\to \cos x = 0, x = \pm \frac{π}{2} + 2 k π \frac{4 k π \pm π}{2}

x = \frac{π}{2} (4 k \pm 1); k \in Z

Answered by mathmax by abdo last updated on 13/Mar/21

e \Rightarrow \frac{1 - \cos (2 x + \frac{π}{2})}{2} - \frac{1 - \cos (2 x - \frac{π}{2})}{2} - \sqrt{7} cosx = 0 \Rightarrow

1 + \sin (2 x) - 1 + \sin (2 x) - 2 \sqrt{7} cosx = 0 \Rightarrow

2 \sin (2 x) - 2 \sqrt{7} cosx = 0 \Rightarrow \sin (2 x) - \sqrt{7} cosx = 0 \Rightarrow

2 sinx cosx - \sqrt{7} cosx = 0 \Rightarrow cosx (2 sinx - \sqrt{7}) = 0 \Rightarrow cosx = 0 or

2 sinx = \sqrt{7}

cosx = 0 \Rightarrow x = \bar{+} \frac{π}{2} + k π and sinx = \frac{\sqrt{7}}{2} is impossible due to \frac{\sqrt{7}}{2} > 1