# Solve the Following Equation: Sec X Cos 5 X + 1 = 0 , 0 < X < π 2 - Mathematics

Sum

Solve the following equation:
$\sec x\cos5x + 1 = 0, 0 < x < \frac{\pi}{2}$

#### Solution

$\sec x\cos5x + 1 = 0$
$\Rightarrow \frac{\cos5x}{\cos x} + 1 = 0$
$\Rightarrow \cos5x + \cos x = 0$
$\Rightarrow 2\cos3x \cos2x = 0$
$\Rightarrow \cos3x = 0 \text{ or } \cos2x = 0$
$\Rightarrow 3x = \left( 2n + 1 \right)\frac{\pi}{2}, n \in Z \text{ or }2x = \left( 2m + 1 \right)\frac{\pi}{2}, m \in Z$
$\Rightarrow x = \left( 2n + 1 \right)\frac{\pi}{6} or x = \left( 2m + 1 \right)\frac{\pi}{4}$
Putting n = 0 and m = 0, we get
$x = \frac{\pi}{6}, \frac{\pi}{4} \left( 0 < x < \frac{\pi}{2} \right)$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 11 Trigonometric equations
Exercise 11.1 | Q 7.3 | Page 22