Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 44920 by peter frank last updated on 06/Oct/18

$$\mathbf{solve}.3^{\mathbf{sin}2\mathbf{x}+2{\mathbf{cos}}^2\mathbf{x}}+3^{1-\mathbf{sin}2\mathbf{x}+2{\mathbf{sin}}^2\mathbf{x}}=28$$

Answered by ajfour last updated on 06/Oct/18

$$3^{\sin 2x+2\cos {}^{2}x}+3^{1-\sin 2x+2\sin {}^{2}x}=28$$$$max(\sin 2x+2\cos {}^{2}x)$$$$=max(\sin 2x+\cos 2x+1)$$$$=\sqrt{2}+1>2$$$$max(1-\sin 2x+2\sin {}^{2}x)$$$$=max(1-\sin 2x+1-\cos 2x)$$$$=2+\sqrt{2}>3$$$$28=3^0+3^3$$$$thisiscertainlypossiblefor$$$$\cos 2x=-1$$$$\Rightarrow 2x=2n\pi +\pi$$$$\Rightarrow x=n\pi +\frac{\pi }{2}\mathrm{\forall }x\in \mathbb{Z}.$$$$$$

Answered by hknkrc46 last updated on 24/Dec/18

$$3^{\sin 2x+2\cos {}^{2}x}+3^{1-\sin 2x+2\sin {}^{2}x}=28$$$$$$$$3^{\sin 2x+(\cos 2x+1)}+3^{1-\sin 2x+(1-\cos 2x)}$$$$3^{(\sin 2x+\cos 2x)+1}+3^{2-(\sin 2x+\cos 2x)}$$$$\Rightarrow \sin 2x+\cos 2x=-1$$$$\Rightarrow 2\sin x\cos x+\cos {}^{2}x-\sin {}^{2}x=-\cos {}^{2}x-\sin {}^{2}x$$$$\Rightarrow 2\cos {}^{2}x+2\sin x\cos x=0$$$$\Rightarrow \cos x(\cos x+\sin x)=0$$$$\Rightarrow \cos x=0\vee \cos x+\sin x=0$$$$\Rightarrow \cos x=0\vee {(\cos x+\sin x)}^2=0$$$$\Rightarrow \cos x=0\vee 1+\sin 2x=0$$$$\Rightarrow \cos x=0\vee \sin 2x=-1$$$$\Rightarrow x=\left(\frac{4k+3}{4}\right)\pi ,k\in \mathbb{Z}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com