Solve. Give the answers in degrees so that 0° ≤ θ < 360°.
1) (5 cos θ - 1) / 2 = 1
answers: 53°, 127°, 233°
2) 3 cot (2θ) + 3 = 0
3) tan^2 θ + 2 sec^2 θ = 3
is my answer for #1 correct and i'm not sure how to do #2 and #3.
3 answers
is the answer to #3 39°?
#3 trick is to get sin^2 + cos^2 = 1 in there
sin^2/cos^2 + 2 /cos^2 = 3
sin^2 + 2 = 3 cos^2
sin^2 + cos^2 + 2 = 4 cos^2
1 + 2 = 4 cos^2
cos^2 = .75
cos = .866
then theta = 30 degrees
sin^2/cos^2 + 2 /cos^2 = 3
sin^2 + 2 = 3 cos^2
sin^2 + cos^2 + 2 = 4 cos^2
1 + 2 = 4 cos^2
cos^2 = .75
cos = .866
then theta = 30 degrees
#2
3 cot (2θ) + 3 = 0
cot(2Ø) = -1
tan(2Ø)= -1
we know tan is negative in II or IV
the reference angle is 45°
that is, we know tan 45° = +1
so 2Ø = 180-45 or 2Ø = 360-45
2Ø = 135° or 2Ø = 315°
Ø = 67.5° or Ø = 157.5°
we also know that the period of tan 2Ø is 90°
so we can add/subtract 90° to get more answers
Ø = 157.5+90 = 247.5
Ø = 247.5+90 = 337.5°
As you can see you will have 4 anwers.
3 cot (2θ) + 3 = 0
cot(2Ø) = -1
tan(2Ø)= -1
we know tan is negative in II or IV
the reference angle is 45°
that is, we know tan 45° = +1
so 2Ø = 180-45 or 2Ø = 360-45
2Ø = 135° or 2Ø = 315°
Ø = 67.5° or Ø = 157.5°
we also know that the period of tan 2Ø is 90°
so we can add/subtract 90° to get more answers
Ø = 157.5+90 = 247.5
Ø = 247.5+90 = 337.5°
As you can see you will have 4 anwers.
1 answer