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Question Number 92585 by john santu last updated on 08/May/20

$$\sin 2\mathrm{z}+5(\sin \mathrm{z}+\cos \mathrm{z})+1=0$$

Commented by john santu last updated on 08/May/20

$$1+\sin 2\mathrm{z}={(\sin \mathrm{z}+\cos \mathrm{z})}^2$$$$\Rightarrow (\sin \mathrm{z}+\cos \mathrm{z})(\sin \mathrm{z}+\cos \mathrm{z}+5)=0$$$$\{\begin{array}{l}\sin \mathrm{z}+\cos \mathrm{z}=0\\ \sin \mathrm{z}+\cos \mathrm{z}=-5\end{array}$$$$\sin \mathrm{z}+\cos \mathrm{z}=0$$$$\left(1\right)\sin \mathrm{z}=-\cos \mathrm{z}\Rightarrow \mathrm{z}=\frac{3\pi }{4}+\pi \mathrm{n}$$$$\left(2\right)\sin \mathrm{z}+\cos \mathrm{z}=0\mid \times \frac{\sqrt{2}}{2}$$$$\sin (\mathrm{z}+\frac{\pi }{4})=0\Rightarrow \mathrm{z}=-\frac{\pi }{4}+\pi \mathrm{n}$$$$\left(3\right)\cos (\frac{\pi }{2}-\mathrm{z})+\cos \mathrm{z}=0$$$$\cos (\mathrm{z}-\frac{\pi }{4})=0\Rightarrow \mathrm{z}=\frac{\pi }{4}\pm \frac{\pi }{2}+\mathrm{n}\pi$$$$$$

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