What is the general solution for sec(2x)+1=0 ?

The general solution for sec(2x)+1=0 is x= pi/2+pin

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sec(2x)+1=0

sec (2 x) + 1 = 0

x = \frac{π}{2} + πn

Degrees

x = 9 0^{\circ} + 18 0^{\circ} n

Solution steps

sec (2 x) + 1 = 0

Move

1

to the right side

sec (2 x) + 1 = 0

Subtract

1

from both sides

sec (2 x) + 1 - 1 = 0 - 1

Simplify

sec (2 x) = - 1

sec (2 x) = - 1

General solutions for

sec (2 x) = - 1

sec (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

2 x = π + 2 πn

2 x = π + 2 πn

Solve

2 x = π + 2 πn : x = \frac{π}{2} + πn

2 x = π + 2 πn

Divide both sides by

2

2 x = π + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{π}{2} + \frac{2 πn}{2}

Simplify

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

sin (θ) = - \frac{4}{7}

6 cos (6 x) = 5

4 sec (x) + 23 = sec (x)

sec (x) sin (x) = 2

sec (x) cot (x) = \frac{7}{6}

What is the general solution for sec(2x)+1=0 ?
The general solution for sec(2x)+1=0 is x= pi/2+pin