Question
Solve the following equations:
$\sec\text{x}\cos5\text{x}+1=0,0<\text{x}<\frac{\pi}{2}$
$\sec\text{x}\cos5\text{x}+1=0,0<\text{x}<\frac{\pi}{2}$
✓
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 $2\text{x}=\frac{\pi}{2}$
$\text{x}=\frac{\pi}{4},\frac{\pi}{6}$
$\frac{\cos5\text{x}+\cos\text{x}}{\cos\text{x}}=0\Rightarrow\cos\text{x}\neq0$
$2\cos3\text{x}\cos2\text{x}=0$
$\cos3\text{x}=0$ 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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