what is sec(3θ) + 5 = -3sec(3θ) - 3?

Question

any work would be greatly appreciated. or an explanation of you you get the answer. i am stuck near the end of the problem because i got to sec(3θ) = 0 and now i dont know what to do

Answer 1

The trig equations with nθ are tough because you have more answers than you think! Let me show you something that helps at the end. But first, we will combine like terms, solve for sec(3θ) and then take the reciprocal of both sides of the equation to turn it into cosine:

4sec(3θ) = -8
sec(3θ) = -2
cos(3θ) = -1/2

At this point you ask yourself "what angles have a cosine of -1/2?" In radians, we have 2π/3 and 4π/3. ALL possible solutions are then 2π/3 + 2πk or 4π/3 + 2πk, where k is an integer. NOW, you set these expressions equal to 3θ and solve for θ:

3θ = 2π/3 + 2πk
θ = 2π/9 + 2πk/3

or

3θ = 4π/3 + 2πk
θ = 4π/9 + 2πk/3

If the question asks you to give all solutions, these are them. If the question asks for all solutions between 0 and 2π, then you start by plugging in k = 0, 1, 2 to get:


θ = 2π/9, 4π/9, 8π/9, 10π/9, 14π/9, 16π/9

Answer 2

sec(3θ) + 5 = -3sec(3θ) - 3
sec(3θ) + 3sec(3θ) = -3 -5
4sec(3θ)=-8
sec(3θ) = 1/cos(3θ)
so
1/cos(3θ) = -2
cos(3θ) = -0.5
so
arccos(3θ) =( the angle whose cos is -.5) = (3θ)
3θ = 120º {you can add any number of whole cycles so}
3θ = 120 + 360n {n=0,1,3.…)
θ =40º, 160º or 280º in the first cycle
or θ(rad) = θº*pi/180 radians = 2Pi/9, 8Pi/9 or 14Pi/9

This is wrong because it only considers one point where cos(3θ) = -0.5 in the first cycle -- there are 2, 120º apart.

Answer 3

sec(3θ) + 5 = -3sec(3θ) - 3

4sec(3θ) + 5 = -3

4sec(3θ) = -8

sec(3θ) = -2

cos3θ = - ½

3θ = 120, 240, 480,...............depending on the range given

θ = 40, 80, 160,......

Answer 4

4sec(3@)=-8
so sec(3@)= -2 and cos3 @=-1/2
so 3@=2pi/3+2kpi
and 3@=4pi/3+2kpi
so
@=2pi/9+2kpi/3
and
@=4pi/9+2kpi/3
Each gives you three different values in the interval 0-2pi
for
k= 0,1,2

Answer 5

θ=120 ℕ - 40

Answer 6

from this equation you get:
4sec(3t) +8 = 0
sect(3t) = -2
t = {sec^-1 (-2) }/ 3