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Question Number 140378 by benjo_mathlover last updated on 07/May/21

$$\sqrt{\sin {}^4\mathrm{x}+4\cos {}^2\mathrm{x}}-\sqrt{\cos {}^4\mathrm{x}+4\sin {}^2\mathrm{x}}=\frac{1}{2}$$$$\mathrm{for}\mathrm{x}\in [0,2\pi ]$$

Commented by john_santu last updated on 07/May/21

$$\sqrt{\sin {}^{4}x+4\cos {}^{2}x}>\sqrt{\cos {}^{4}x+4\sin {}^{2}x}$$$$\Rightarrow \sin {}^{2}x(\sin {}^{2}x-4)>\cos {}^{2}x(\cos {}^{2}x-4)$$$$\Rightarrow \sin {}^{2}x>\cos {}^{2}x$$$$\Rightarrow \sin {}^{2}x>1-\sin {}^{2}x$$$$(\sqrt{2}\sin x-1)(\sqrt{2}\sin x+1)>0$$$$\Rightarrow \sin x<-\frac{1}{\sqrt{2}}\cup \sin x>\frac{1}{\sqrt{2}}$$

Answered by john_santu last updated on 07/May/21

$$\sqrt{\sin {}^{4}x+4-4\sin {}^{2}x}-\sqrt{\cos {}^{4}x+4-4\cos {}^{2}x}=\frac{1}{2}$$$$\sqrt{{(2-\sin {}^{2}x)}^2}-\sqrt{{(2-\cos {}^{2}x)}^2}=\frac{1}{2}$$$$\cancel{2}-\sin {}^{2}x+\cos {}^{2}x-\cancel{2}=\frac{1}{2}$$$$\cos 2x=\frac{1}{2}$$$$\cos 2x=\cos \frac{\pi }{3}$$$$2x=\{\begin{array}{l}-\frac{\pi }{3}+2n\pi \\ \frac{\pi }{3}+2n\pi \end{array}\iff \{\begin{array}{l}x=-\frac{\pi }{6}+n\pi \\ x=\frac{\pi }{6}+n\pi \end{array}$$$$x=\{\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{3},\frac{11\pi }{6}\}$$

Answered by MJS_new last updated on 07/May/21

$$\sqrt{{\sin }^{4}x+{4\cos }^{2}x}-\sqrt{{\cos }^{4}x+{4\sin }^{2}x}=\frac{1}{2}$$$$\mathrm{let}t=\tan x$$$$\sqrt{{\left(\frac{t}{\sqrt{t^2+1}}\right)}^4+4{\left(\frac{1}{\sqrt{t^2+1}}\right)}^2}-\sqrt{{\left(\frac{1}{\sqrt{t^2+1}}\right)}^4+4{\left(\frac{t}{\sqrt{t^2+1}}\right)}^2}=\frac{1}{2}$$$$\sqrt{{\left(\frac{t^2+2}{t^2+1}\right)}^2}-\sqrt{{\left(\frac{2t^2+1}{t^2+1}\right)}^2}=\frac{1}{2}$$$$\frac{1-t^2}{1+t^2}=\frac{1}{2}$$$$\Rightarrow t=\pm \frac{\sqrt{3}}{3}=\tan x$$$$\Rightarrow x=\frac{\pi }{6}\vee x=\frac{5\pi }{6}\vee x=\frac{7\pi }{6}\vee x=\frac{11\pi }{6}$$

Commented by greg_ed last updated on 07/May/21

$$\mathrm{right}!$$

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