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Question Number 149478 by mathdanisur last updated on 05/Aug/21

$$Solveforrealnumbers:$$$$x^2-2x-3sin\mathit{y}-4cos\mathit{y}+6=0$$

Commented by iloveisrael last updated on 06/Aug/21

$$\mathrm{\Delta }⩾0$$$$3\sin \mathrm{y}+4\mathrm{cosy}-6⩾-1$$$$5\sin (\mathrm{y}+{\tan }^{-1}\left(\frac{4}{3}\right))⩾5$$$$\sin (\mathrm{y}+\tan {}^{-1}\left(\frac{4}{3}\right))=1$$$$\Rightarrow \mathrm{y}+{\tan }^{-1}\left(\frac{4}{3}\right)=\frac{\pi }{2}+2\mathrm{k}\pi$$$$\Rightarrow \mathrm{y}=\frac{\pi }{2}-{\tan }^{-1}\left(\frac{4}{3}\right)+2\mathrm{k}\pi$$

Commented by mathdanisur last updated on 06/Aug/21

$$\mathrm{Thankyou}\mathbf{Ser}$$

Answered by mr W last updated on 05/Aug/21

$$3sin\mathit{y}-4cos\mathit{y}=x^2-2x+6$$$$5(sin\mathit{y}\frac{3}{5}-cos\mathit{y}\frac{4}{5})=x^2-2x+6$$$$5(sin\mathit{y}\cos \alpha -cos\mathit{y}\sin \alpha )={(x-1)}^2+5$$$$with\cos \alpha =\frac{3}{5},\sin \alpha =\frac{4}{5},\tan \alpha =\frac{4}{3}$$$$5\sin (y-\alpha )={(x-1)}^2+5$$$$5\sin (y-{\tan }^{-1}\frac{4}{3})={(x-1)}^2+5$$$$LHS⩽5$$$$RHS⩾5$$$$\Rightarrow LHS=RHS=5$$$$\Rightarrow x=1$$$$\Rightarrow \sin (y-{\tan }^{-1}\frac{4}{3})=1$$$$\Rightarrow y-{\tan }^{-1}\frac{4}{3}=2k\pi +\frac{\pi }{2}$$$$\Rightarrow y=2k\pi +\frac{\pi }{2}+{\tan }^{-1}\frac{4}{3}$$

Commented by mathdanisur last updated on 05/Aug/21

$$\mathrm{Thank}\mathrm{you}\mathbf{Ser},\mathrm{cool}$$

Commented by peter frank last updated on 05/Aug/21

$$explanation2ndand3rdline$$

Commented by mr W last updated on 05/Aug/21

$$iaddedsomelinesmore.$$

Commented by peter frank last updated on 05/Aug/21

$$thankyou$$

Commented by Tawa11 last updated on 05/Aug/21

$$\mathrm{great}$$

Answered by MJS_new last updated on 05/Aug/21

$$x=1\pm \sqrt{-5+3\sin y+4\cos y}$$$$-10⩽-5+3\sin y+4\cos y⩽0$$$$\Rightarrow$$$$-5+3\sin y+4\cos y=0\iff y=2n\pi +\arctan \frac{3}{4}$$$$\Rightarrow$$$$\mathrm{solution}\mathrm{is}$$$$x=1\wedge y=2n\pi +\arctan \frac{3}{4}\mathrm{\forall }n\in \mathbb{Z}$$

Commented by mathdanisur last updated on 05/Aug/21

$$\mathrm{Thank}\mathrm{you}\mathbf{Ser},\mathrm{cool}$$

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