Question

Solve the equation: sec x . cos 5x + 1 = 0; 0 < x < 2π.

Answer 1

Answer:   x = π/6 ,  x = π/2  , x = π/4 AND  x = 3π/4    for  0 < x < 2π

Explanation:

Given that

sec x . cos 5x + 1 = 0       ;      0 < x < 2π

cos 5x / cos x + 1 = 0

Multiplying both sides by cos x , we get

cos 5x + cos x + = 0

2cos[(5x + x)/2].cos[(5x - x)/2] = 0

cos3x . cos2x = 0/2

cos3x . cos2x = 0

cos3x = 0     ;    cos2x = 0

For cos3x = 0 :

cos3x = 0

3x = π/2 AND  3x = 3π/2

x = π/6 AND  x = π/2

AND

For cos3x = 0 :

2x = π/2 AND  2x = 3π/2

x = π/4 AND  x = 3π/4

Answer 2

Step-by-step explanation:

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