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

$$\mathrm{Solve}\mathrm{for}\mathrm{real}\mathrm{numbers}:$$$$\frac{1}{{\sin }^{2\mathbf{k}}\left(\mathrm{x}\right)}+\frac{1}{{\cos }^{2\mathbf{k}}\left(\mathrm{x}\right)}=8;\mathrm{k}\in \mathbb{Z}$$

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

$$\text{Missing left or extra right}$$$$thatmeansfork⩾3,LHS⩾16\Rightarrow nosolutionforLHS=8!$$$$fork=0:1+1=8\Rightarrow nosolution!$$$$fork⩽-1:LHS⩽2\Rightarrow nosolutionforLHS=8!$$$$thereforetheonlypossibilityisk=1or2.$$$$$$$$k=1:$$$$\frac{1}{{\sin }^{2}x}+\frac{1}{{\cos }^{2}x}=8$$$$\frac{1}{{\sin }^{2}x{\cos }^{2}x}=8$$$$\frac{1}{{\sin }^{2}2x}=2$$$$\Rightarrow \sin 2x=\pm \frac{1}{\sqrt{2}}$$$$\Rightarrow 2x=n\pi \pm \frac{\pi }{4}$$$$\Rightarrow x=\frac{(2n+1)\pi }{8}fork=1$$$$$$$$k=2:$$$$\frac{1}{{\sin }^{4}x}+\frac{1}{{\cos }^{4}x}=8$$$$\frac{{\sin }^{4}x+{\cos }^{4}x}{{\left(\sin x\cos x\right)}^4}=8$$$$\frac{(1-\frac{{\sin }^{2}2x}{2})16}{{\sin }^{4}2x}=8$$$$\frac{(2-\frac{{\sin }^{2}2x}{1})}{{\sin }^{4}2x}=1$$$${\sin }^{2}2x+{\sin }^{2}2x-2=0$$$$({\sin }^{2}2x+2)({\sin }^{2}2x-1)=0$$$$\sin 2x=\pm 1$$$$2x=\frac{(2n+1)\pi }{2}$$$$\Rightarrow x=\frac{(2n+1)\pi }{4}fork=2$$

Commented by MathSh last updated on 21/Oct/21

$$\mathrm{Perfect}\mathrm{dear}\mathrm{Ser},\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}$$

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