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.
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
#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.
To solve equation 1:
(5 cos θ - 1) / 2 = 1
First, multiply both sides of the equation by 2 to isolate 5 cos θ - 1 on the left side:
5 cos θ - 1 = 2
Next, add 1 to both sides of the equation:
5 cos θ = 3
Now, divide both sides of the equation by 5:
cos θ = 3/5
To find the possible solutions for θ, we can use the inverse cosine function (cos⁻¹) and plug in the value 3/5:
θ = cos⁻¹(3/5)
Using a calculator or a trigonometric table, we find
θ ≈ 53°, 127°, 233°
So, your answers of 53°, 127°, and 233° are correct.
To solve equation 2:
3 cot (2θ) + 3 = 0
Let's start by subtracting 3 from both sides of the equation:
3 cot (2θ) = -3
Next, divide both sides of the equation by 3:
cot (2θ) = -1
Taking the reciprocal of both sides gives:
tan (2θ) = -1
Now, we need to find the possible angles whose tangent is -1. The tangent function is negative in the second and fourth quadrants. We can express -1 as tan(-π/4) or tan(3π/4). Therefore:
2θ = -π/4 or 2θ = 3π/4
Dividing both sides by 2 gives:
θ = -π/8 or θ = 3π/8
To convert the angles to degrees, multiply by 180/π:
θ ≈ -22.5° or θ ≈ 67.5°
So, the solutions for θ in equation 2 are approximately -22.5° and 67.5°.
To solve equation 3:
tan^2 θ + 2 sec^2 θ = 3
First, we can rewrite sec^2 θ as 1 + tan^2 θ:
tan^2 θ + 2(1 + tan^2 θ) = 3
Simplifying the equation gives:
tan^2 θ + 2 + 2 tan^2 θ = 3
Combining like terms:
3 tan^2 θ + 2 = 3
Next, subtract 2 from both sides of the equation:
3 tan^2 θ = 1
Dividing both sides by 3 gives:
tan^2 θ = 1/3
Now, take the square root of both sides to find tan θ:
tan θ = ±√(1/3)
To find the possible solutions, we use the inverse tangent function (tan⁻¹):
θ = tan⁻¹(±√(1/3))
Using a calculator or a trigonometric table, we find:
θ ≈ ±30°, ±150°
So, the solutions for θ in equation 3 are approximately ±30° and ±150°.
In summary:
1) The answers for equation 1 are 53°, 127°, and 233°.
2) The solutions for equation 2 are approximately -22.5° and 67.5°.
3) The solutions for equation 3 are approximately ±30° and ±150°.