Solve-for-real-numbers-1-sin-2k-x-1-cos-2k-x-8-k-Z-

Question Number 157258 by MathSh last updated on 21/Oct/21

Solve for real numbers :

\frac{1}{\sin^{2 k} (x)} + \frac{1}{\cos^{2 k} (x)} = 8; k \in Z

Commented by mr W last updated on 21/Oct/21

Missing \left or extra \right

t h a t m e a n s f o r k ⩾ 3, L H S ⩾ 16 \Rightarrow n o s o l u t i o n f o r L H S = 8!

f o r k = 0 : 1 + 1 = 8 \Rightarrow n o s o l u t i o n!

f o r k ⩽ - 1 : L H S ⩽ 2 \Rightarrow n o s o l u t i o n f o r L H S = 8!

t h e r e f o r e t h e o n l y p o s s i b i l i t y i s k = 1 o r 2 .

k = 1 :

\frac{1}{\sin^{2} x} + \frac{1}{\cos^{2} x} = 8

\frac{1}{\sin^{2} x \cos^{2} x} = 8

\frac{1}{\sin^{2} 2 x} = 2

\Rightarrow \sin 2 x = \pm \frac{1}{\sqrt{2}}

\Rightarrow 2 x = n π \pm \frac{π}{4}

\Rightarrow x = \frac{(2 n + 1) π}{8} f o r k = 1

k = 2 :

\frac{1}{\sin^{4} x} + \frac{1}{\cos^{4} x} = 8

\frac{\sin^{4} x + \cos^{4} x}{{(\sin x \cos x)}^{4}} = 8

\frac{(1 - \frac{\sin^{2} 2 x}{2}) 16}{\sin^{4} 2 x} = 8

\frac{(2 - \frac{\sin^{2} 2 x}{1})}{\sin^{4} 2 x} = 1

\sin^{2} 2 x + \sin^{2} 2 x - 2 = 0

(\sin^{2} 2 x + 2) (\sin^{2} 2 x - 1) = 0

\sin 2 x = \pm 1

2 x = \frac{(2 n + 1) π}{2}

\Rightarrow x = \frac{(2 n + 1) π}{4} f o r k = 2

Commented by MathSh last updated on 21/Oct/21

Perfect dear Ser, thank you so much

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