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Question Number 145953 by mathdanisur last updated on 09/Jul/21

Commented by mathdanisur last updated on 09/Jul/21

$$Solvethetrigonometricequation$$

Commented by puissant last updated on 09/Jul/21

$$\mathrm{What}\mathrm{do}\mathrm{you}\mathrm{need}?$$

Commented by iloveisrael last updated on 10/Jul/21

Commented by mr W last updated on 10/Jul/21

$$yoursolutionsare:$$$$x=k\pi \pm \frac{3\pi }{8},x=k\pi \pm \frac{\pi }{8}$$$$thatmeans:$$$$x=...,-\frac{3\pi }{8},-\frac{\pi }{8},\frac{\pi }{8},\frac{3\pi }{8},\frac{5\pi }{8},\frac{7\pi }{8},...$$$$thiscanalsobewrittenas$$$$x=\frac{k\pi }{4}+\frac{\pi }{8}$$

Answered by mr W last updated on 09/Jul/21

$${({\sin }^{4}x+{\cos }^{4}x)}^2-2{\sin }^{4}x{\cos }^{4}x=\frac{17}{16}{\cos }^2\left(2x\right)$$$${({({\sin }^{2}x+{\cos }^{2}x)}^2-2{\sin }^{2}x{\cos }^{2}x)}^2-\frac{{\sin }^4\left(2x\right)}{8}=\frac{17}{16}{\cos }^2\left(2x\right)$$$$16{(1-\frac{{\sin }^{2}2x}{2})}^2-2{\sin }^4\left(2x\right)=17(1-{\sin }^2\left(2x\right))$$$$4{(2-{\sin }^{2}2x)}^2-2{\sin }^4\left(2x\right)=17(1-{\sin }^2\left(2x\right))$$$$lett=\sin 2x$$$$4{(2-t^2)}^2-2t^4=17(1-t^2)$$$$2t^4+t^2-1=0$$$$(2t^2-1)(t^2+1)=0$$$$t^2=\frac{1}{2}$$$$\Rightarrow t=\sin 2x=\pm \frac{1}{\sqrt{2}}$$$$\Rightarrow 2x=\frac{k\pi }{2}+\frac{\pi }{4}$$$$\Rightarrow x=\frac{k\pi }{4}+\frac{\pi }{8}$$

Commented by mr W last updated on 10/Jul/21

$$from\sin 2x=\pm \frac{1}{\sqrt{2}},weget$$$$2x=k\pi \pm \frac{\pi }{4}(k\in Z)$$$$thisisthesameas2x=\frac{k\pi }{2}+\frac{\pi }{4}(k\in Z)$$$$bothmean2x=...,-\frac{\pi }{4},\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},etc.$$

Commented by mathdanisur last updated on 09/Jul/21

$$CoolSer,thankyou$$

Commented by iloveisrael last updated on 10/Jul/21

$$\mathrm{why}\sin 2\mathrm{x}=\pm \frac{1}{\sqrt{2}}\mathrm{then}$$$$2\mathrm{x}=\frac{\mathrm{k}\pi }{2}+\frac{\pi }{4}?\mathrm{not}2\mathrm{x}=2\mathrm{k}\pi +\frac{\pi }{4}$$$$?$$

Commented by iloveisrael last updated on 10/Jul/21

$$\mathrm{what}\mathrm{periodic}\mathrm{of}\sin \mathrm{function}$$$$2\pi \mathrm{or}\pi$$$$\mathrm{it}\mathrm{should}\mathrm{be}2\mathrm{x}=2\mathrm{k}\pi \pm \frac{\pi }{4}$$$$\mathrm{then}\mathrm{x}=\mathrm{k}\pi \pm \frac{\pi }{8}$$

Commented by mr W last updated on 10/Jul/21

$$wehave\sin 2x=\frac{1}{\sqrt{2}}and\sin 2x=-\frac{1}{\sqrt{2}}!$$$$from\sin 2x=\frac{1}{\sqrt{2}}weget$$$$2x=2k\pi +\frac{\pi }{4}or2x=(2k+1)\pi -\frac{\pi }{4}.$$$$from\sin 2x=-\frac{1}{\sqrt{2}}weget$$$$2x=2k\pi -\frac{\pi }{4}or2x=(2k+1)\pi +\frac{\pi }{4}.$$$$summarized2x=k\pi \pm \frac{\pi }{4}.$$$$$$$$justhavealookatthegraphof$$$$function\sin x.$$

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