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Question Number 63703 by kaivan.ahmadi last updated on 07/Jul/19

$${sin}^{3}x+{cos}^{3}x=1-\frac{1}{2}sin2x:x\in [0,2\pi ].$$$$findx$$

Commented by Prithwish sen last updated on 08/Jul/19

$$(\mathrm{sinx}+\mathrm{cosx})(1-\frac{1}{2}\sin 2\mathrm{x})-(1-\frac{1}{2}\sin 2\mathrm{x})=0$$$$\Rightarrow \sin 2\mathrm{x}=2\mathrm{which}\mathrm{is}\mathrm{impossible}$$$$\mathrm{and}\sin (\frac{\pi }{4}+\mathrm{x})=\sin \frac{\pi }{4}\mathrm{or}\sin \frac{3\pi }{4}$$

Commented by kaivan.ahmadi last updated on 08/Jul/19

$$(sinx+cosx-1)(1-\frac{1}{2}sin2x)=0\Rightarrow$$$$sinx+cosx=1\Rightarrow$$$$x=0,\frac{\pi }{2},2\pi$$

Answered by MJS last updated on 07/Jul/19

$$x=2\arctan t$$$$-\frac{2t(t-1)(t^4+2t^3+2t^2-2t+1)}{{(t^2+1)}^3}=0$$$$t_1=0$$$$t_2=1$$$$\mathrm{no}\mathrm{other}\mathrm{real}\mathrm{solutions}$$$$\Rightarrow x=0\vee x=\frac{\pi }{2}$$

Commented by kaivan.ahmadi last updated on 08/Jul/19

$$1.\frac{5\pi }{2}$$$$2.\frac{7\pi }{2}$$$$3.2\pi$$$$4.3\pi$$$$excusemesirfindsumofanswers$$

Commented by kaivan.ahmadi last updated on 08/Jul/19

$$x=2\pi isananswer$$

Commented by MJS last updated on 08/Jul/19

$$\mathrm{generally}x=2n\pi \vee x=\frac{\pi }{2}+2n\pi ;n\in \mathbb{Z}$$

Commented by MJS last updated on 08/Jul/19

$$\frac{7\pi }{2}\mathrm{and}3\pi \mathrm{are}\mathrm{wrong}$$

Commented by kaivan.ahmadi last updated on 08/Jul/19

$$sircanyouexplainhowdoyou$$$$changethevariable?$$

Commented by MJS last updated on 08/Jul/19

$$\mathrm{the}\mathrm{substitution}x=2\arctan t\mathrm{leads}\mathrm{to}$$$$\sin x=\frac{2t}{t^2+1}$$$$\cos x=-\frac{t^2-1}{t^2+1}$$

Commented by kaivan.ahmadi last updated on 08/Jul/19

$$thankssir$$

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