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Trigonometric Equations

Section: Mathematics

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Question : 5 of 24

Marks: +1, -0

The number of solutions of the equation

\sec x \cos 5x + 1 = 0

in the interval

[0,2\pi]

[AP EAPCET 23 May 2025 S1]

5
8
10
12

Solution:

Given, trigonometric equation is

\sec x \cdot \cos 5x + 1 = 0

\;\Rightarrow\;\; \frac{\cos 5x + \cos x}{\cos x}=0

\;\Rightarrow\;\; 2 \cdot \cos 3x \cdot \cos 2x = 0

\;\Rightarrow\;\; \cos 3x = 0 \;\text{ or }\; \cos 2x = 0

\;\Rightarrow\;\; 3x = \frac{\pi}{2},\; \frac{3\pi}{2},\; \frac{5\pi}{2},\; \frac{7\pi}{2},\; \frac{9\pi}{2},\; \frac{11\pi}{2}

\;\; \binom{\because 0\leq x\leq 2\pi}{\therefore 0\leq 3x\leq 6\pi}

\;\Rightarrow\;\; x = \frac{\pi}{6},\; \frac{3\pi}{6},\; \frac{5\pi}{6},\; \frac{7\pi}{6},\; \frac{9\pi}{6},\; \frac{11\pi}{6}

\;\text{ Also, }\; 2x = \frac{\pi}{2},\; \frac{3\pi}{2},\; \frac{5\pi}{2},\; \frac{7\pi}{2} \binom{\because 0\leq x\leq 2\pi}{\therefore 0\leq 2x\leq 4\pi}

\;\Rightarrow\;\; x = \frac{\pi}{4},\; \frac{3\pi}{4},\; \frac{5\pi}{4},\; \frac{7\pi}{4}

∴ Total number of solutions

=10

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