Question

Find the number of solutions satisfying secxcos5x+1=0 where 0<x<pie

Answer 1

Dear student

$$\begin{gathered} \mathrm{secx}\cos 5\mathrm{x}+1=0 \\ \frac{\cos 5\mathrm{x}+\mathrm{cosx}}{\mathrm{cosx}}=0 \\ \frac{2\cos 3\mathrm{x}\cos 2\mathrm{x}}{\mathrm{cosx}}=0 \\ \frac{2\left(4{\cos }^3\mathrm{x}-3\mathrm{cosx}\right)\left(2{\cos }^2\mathrm{x}-1\right)}{\mathrm{cosx}}=0 \\ \frac{4{\cos }^3\mathrm{x}-3\mathrm{cosx}}{\mathrm{cosx}}=0\mathrm{and}2{\cos }^2\mathrm{x}-1=0 \\ 4{\cos }^2\mathrm{x}-3=0\mathrm{and}2{\cos }^2\mathrm{x}-1=0 \\ \mathrm{cosx}=\frac{\sqrt{3}}{2}\mathrm{or}\mathrm{cosx}=-\frac{\sqrt{3}}{2}\mathrm{and}\mathrm{cosx}=\frac{1}{\sqrt{2}}\mathrm{or}\mathrm{cosx}=-\frac{1}{\sqrt{2}} \\ \mathrm{x}=\frac{\mathrm{\pi }}{6},\mathrm{or}\mathrm{x}=\frac{5\mathrm{\pi }}{6}\mathrm{and}\mathrm{x}=\frac{\mathrm{\pi }}{4}\mathrm{or}\mathrm{x}=\frac{3\mathrm{\pi }}{4} \\ \mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{solutions}\mathrm{are}4 \end{gathered}$$
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