Principal solutions of the equation sin 2x + cos 2x = 0, where π < x < 2π are

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$$\begin{gathered} \mathrm{\pi }<\mathrm{x}<2\mathrm{\pi } \\ \Rightarrow 2\mathrm{\pi }<2\mathrm{x}<4\mathrm{\pi } \\ \sin \left(2\mathrm{x}\right)+\cos \left(2\mathrm{x}\right)=0 \\ \Rightarrow \sin \left(2\mathrm{x}\right)=-\cos \left(2\mathrm{x}\right) \\ \Rightarrow \frac{\sin \left(2\mathrm{x}\right)}{\cos \left(2\mathrm{x}\right)}=-1 \\ \Rightarrow \tan \left(2\mathrm{x}\right)=-1 \\ \mathrm{tanx} \mathrm{is} \mathrm{negative} \mathrm{in} \mathrm{second} \mathrm{and} \mathrm{fourth} \mathrm{quadrant}. \\ \mathrm{Also} \\ \tan \left(\frac{3\mathrm{\pi }}{4}\right)=\tan \left(\frac{7\mathrm{\pi }}{4}\right)=\tan \left(\frac{11\mathrm{\pi }}{4}\right)=\tan \left(\frac{15\mathrm{\pi }}{4}\right)=...=-1 \\ \mathrm{But} 2\mathrm{\pi }<2\mathrm{x}<4\mathrm{\pi } \\ 2\mathrm{x}=\frac{11\mathrm{\pi }}{4} \mathrm{or} 2\mathrm{x}=\frac{15\mathrm{\pi }}{4} \\ \mathbf{\Rightarrow }\mathbf{x}\mathbf{=}\frac{\mathbf{11}\mathbf{\pi }}{\mathbf{8}}\mathbf{ }\mathbf{or}\mathbf{ }\mathbf{x}\mathbf{=}\frac{\mathbf{15}\mathbf{\pi }}{\mathbf{8}} \end{gathered}$$

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