Question
Solve the following equations: $\sec\text{x}\cos5\text{x}+1=0,0<\text{x}<\frac{\pi}{2}$

Answer

$\sec\text{x}\cos5\text{x}+1=0$ $\frac{\cos5\text{x}+\cos\text{x}}{\cos\text{x}}=0\Rightarrow\cos\text{x}\neq0$ $2\cos3\text{x}\cos2\text{x}=0$ $\cos3\text{x}=0$ or $\cos2\text{x}=0$ $3\text{x}=\frac{\pi}{2}$ or $2\text{x}=\frac{\pi}{2}$ $\text{x}=\frac{\pi}{4},\frac{\pi}{6}$

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