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Question Number 157265 by MathSh last updated on 21/Oct/21

$$\mathrm{Solve}\mathrm{for}\mathrm{real}\mathrm{numbers}:$$$$\sin \left(\mathrm{x}\right)+\cos \left(\mathrm{x}\right)+\sec \left(\mathrm{x}\right)\cdot \csc \left(\mathrm{x}\right)=2+\sqrt{2}$$

Answered by TheSupreme last updated on 21/Oct/21

$$sin\left(x\right)+cos\left(x\right)-\frac{1}{sin\left(x\right)cos\left(x\right)}=A$$$$\pm s\sqrt{1-s^2}-(\pm \frac{1}{s\sqrt{1-s^2}})=A$$$$s^2\sqrt{1-s^2}+s-s^3=\pm As\sqrt{1-s^2}$$$$(s^2-(\pm As))\sqrt{1-s^2}=s^3-s$$$$s\ne 0$$$$(s-(\pm A))\sqrt{1-s^2}=s^2-1$$$$(s-(\pm A))=-\sqrt{1-s^2}$$$$s^2+A^2-(\pm 2sA)=1-s^2$$$$2s^2-(\pm 2sA)+A^2-1=0$$$$s_{1,2}=\frac{2A\pm \sqrt{4A^2+8A^2}}{4}=A\frac{1\pm \sqrt{3}}{2}\left(a\right)$$$$s_{1,2}=A\frac{-1\pm \sqrt{3}}{2}$$$$...$$$$s⩽1$$$$s_1=(2+\sqrt{2})\left(\frac{1-\sqrt{3}}{2}\right)\left(a\right)$$$$cos=\sqrt{1-s^2}$$$$s_2=(2+\sqrt{2})\left(\frac{-1+\sqrt{3}}{2}\right)\left(b\right)$$$$cos=-\sqrt{1-s^2}$$

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