What is the general solution for 3sec(θ)+3=0 ?

The general solution for 3sec(θ)+3=0 is θ=pi+2pin

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3sec(θ)+3=0

3 sec (θ) + 3 = 0

θ = π + 2 πn

Degrees

θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

3 sec (θ) + 3 = 0

Move

3

to the right side

3 sec (θ) + 3 = 0

Subtract

3

from both sides

3 sec (θ) + 3 - 3 = 0 - 3

Simplify

3 sec (θ) = - 3

3 sec (θ) = - 3

Divide both sides by

3

3 sec (θ) = - 3

Divide both sides by

3

\frac{3 sec ( θ )}{3} = \frac{- 3}{3}

Simplify

sec (θ) = - 1

sec (θ) = - 1

General solutions for

sec (θ) = - 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}

θ = π + 2 πn

θ = π + 2 πn

12 \cdot 9.8 \cdot sin (a) - 0.6 \cdot 12 \cdot 9.8 \cdot cos (a) = 12 \cdot 1.79

sin (x) (sin (x) - 1) = 5 cos^{2} (x) - 4

v, a = arctan (\frac{v ^{2}}{g R})

3 sin^{2} (x) = 4 sin (x)

5 sin (θ) = 4

What is the general solution for 3sec(θ)+3=0 ?
The general solution for 3sec(θ)+3=0 is θ=pi+2pin