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Question Number 179449 by Shrinava last updated on 29/Oct/22

$$\mathrm{Solve}\mathrm{for}\mathrm{real}\mathrm{numbers}:$$$$3\mathrm{sinx}+4(\mathrm{y}+\mathrm{cosx})={\mathrm{y}}^2+9$$

Answered by mr W last updated on 29/Oct/22

$$3\sin x+4\cos x-5=y^2-4y+4$$$$5[\sin (x+\alpha )-1]={(y-2)}^2$$$$since\sin (x+\alpha )⩽1,LHS⩽0.$$$$butRHS⩾0,therefore$$$$LHS=RHS=0$$$$\Rightarrow \sin (x+\alpha )=1\Rightarrow x=2k\pi +\frac{\pi }{2}-{\tan }^{-1}\frac{4}{3}$$$$\Rightarrow y=2$$

Commented by Shrinava last updated on 01/Nov/22

$$\mathrm{cool}\mathrm{dear}\mathrm{professor}\mathrm{thank}\mathrm{you}$$

Answered by manxsol last updated on 29/Oct/22

$$3sinx+4cosx=y^2-4y+9$$$$\frac{3}{5}sinx+\frac{4}{5}cosx=\frac{y^2-4y+9}{5}$$$$sin(x+\theta )=\frac{y^2-4y+9}{5}$$$$\theta =arsen\left(3/5\right)$$$$-1⩽\frac{y^2-4y+9}{5}⩽1$$$$-5⩽y^2-4y+9\hat{\mathrm{\Lambda }}{y}^2-4y+9⩽5$$$$0⩽y^2-4y+14\mathrm{\Delta }=16-56$$$$\mathrm{\Delta }<0\Rightarrow y^2-4y+14⟩0\mathrm{\forall }y>$$$$y^2-4y+9⩽5$$$$y^2-4y+4⩽0$$$${(y-2)}^2⩽0$$$$solutiony=2$$$$sin(x+\theta )=\frac{2^2-4\left(2\right)+9}{5}=1$$$$x+\theta =\frac{\pi }{2}+2k\pi$$$$x=(\frac{\pi }{2}-arcsin\left(\frac{3}{5}\right))+2k\pi$$$$$$$$$$$$$$$$$$

Commented by Shrinava last updated on 01/Nov/22

$$\mathrm{cool}\mathrm{dear}\mathrm{professor}\mathrm{thank}\mathrm{you}$$

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