What is the general solution for 3sec(θ)+1=7,0<= θ<2pi ?

The general solution for 3sec(θ)+1=7,0<= θ<2pi is θ= pi/3 ,θ=(5pi)/3

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3sec(θ)+1=7,0<= θ<2pi

3 sec (θ) + 1 = 7, 0 \leq θ < 2 π

θ = \frac{π}{3}, θ = \frac{5 π}{3}

Degrees

θ = 6 0^{\circ}, θ = 30 0^{\circ}

Solution steps

3 sec (θ) + 1 = 7, 0 \leq θ < 2 π

Move

1

to the right side

3 sec (θ) + 1 = 7

Subtract

1

from both sides

3 sec (θ) + 1 - 1 = 7 - 1

Simplify

3 sec (θ) = 6

3 sec (θ) = 6

Divide both sides by

3

3 sec (θ) = 6

Divide both sides by

3

\frac{3 sec ( θ )}{3} = \frac{6}{3}

Simplify

sec (θ) = 2

sec (θ) = 2

General solutions for

sec (θ) = 2

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 = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Solutions for the range

0 \leq θ < 2 π

θ = \frac{π}{3}, θ = \frac{5 π}{3}

sin (2 x) = 5 sin (x)

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

- 2 sin^{2} (x) - sin (x) + 3 = 0

tan (x) = - 3 + 2

3 tan (x) - cot (x) = 0

What is the general solution for 3sec(θ)+1=7,0<= θ<2pi ?
The general solution for 3sec(θ)+1=7,0<= θ<2pi is θ= pi/3 ,θ=(5pi)/3