Question

The number of solution of $\sec x\cos 5x+1=0$ in the interval $[0,2\pi ]$ is

• A. $5$
• B. $8$
• C. $10$
• D. $12$

Answer 1

The correct option is B $8$

• Given- sec(x)cos(5x)+1=0
  
• cos 5x/cos x + 1 = 0
• cos 5x + cos x = 0
• 2(cos 3x )( cos 2x) = 0 (cosA+cosB=2cos((A+B)/2)cos((A-B)/2)
• (i) cos 3x = 0 , 3x = (2 n + 1)π/2 , x = (2n + 1 ) π/6
• (ii)cos 2x = 0, 2x = (2 n + 1)π/2 , x = (2 n + 1)π/4
  

Therefore in [0,2π]

x= π/6, 5π/6, 7π/6, 11π/6, π/4, 3π/4, 5π/4, 7π/4

sec⁡xcos⁡5x+1=cos 5x/cos x + 1 = 0 cos 5x + cos x = 0 2cos 3x * cos 2x = 0 (i) cos 3x = 0 3x = (2 n + 1)π/2 x = (2n + 1 ) π/6 (ii)cos 2x = 0 2x = (2 n + 1)π/2 x = (2 n + 1)π/4