Christopher J. answered • 05/25/20

Berkeley Grad Math Tutor (algebra to calculus)

Eric:

1) Yes x=-33.7 is indeed an answer. Because we are in the interval 0 ≤ x ≤ 360, x = 360-33.7 = 326.3

Remember, we are finding where tan(x) = -2/3. tan is negative in the 2nd and 4th quadrants. The second quadrant corresponds to 180-33.7 = 146.3.

2) I'll let you try to figure this one out.

3) ((cosθ-4)/sinθ)-(4sinθ)/(5cosθ-2)=0

We put the expression under the same denominator

[(cos(θ)-4)*(5cos(θ)-2) - 4sin(θ)*sin(θ)] / [sin(θ)*(5cos(θ)-2)] = 0

We really only care about the numerator being equal to 0.

(cos(θ)-4)*(5cos(θ)-2)-4sin(θ)*sin(θ) = 0

5cos^{2}(θ) - 2cos(θ) -20cos(θ) + 8 - 4sin^{2}(θ) = 0 (using FOIL)

5cos^{2}(θ) -22cos(θ) + 8 - 4sin^{2}(θ) = 0

5cos^{2}(θ) -22cos(θ) +8 -4(1-cos^{2}(θ)) = 0 (remember sin^{2}(θ) = 1-cos^{2}(θ))

5cos^{2}(θ) -22cos(θ) +8 -4 +4cos^{2}(θ) = 0

9cos^{2}(θ) -22cos(θ) +4 = 0

4)

We solve (1/cos^{2}(θ)) = 3/sin(θ)

sin(θ) = 3cos^{2}(θ)

sin(θ) = 3(1-sin^{2}(θ))

sin(θ) = 3 - 3sin^{2}(θ)

3sin^{2}(θ) +sin(θ) - 3 = 0

let u =sin(θ)

solve 3u^{2}+u-3 = 0

u= -1.18 or u = 0.847

We reject u = -1.18 because -1≤sin(θ)≤1

So we need to solve sin(θ) = 0.847 . I'll let you solve this equation. Hint there are two answers.

5)

(1-sinθ)/(1+sinθ) = [(1-sin(θ))/(1+sin(θ))] * [(1-sin(θ))/(1-sin(θ)]

(1-sinθ)/(1+sinθ) = (1-sin(θ))^{2} / [(1+sin(θ))*(1-sin(θ))]

(1-sinθ)/(1+sinθ) = (1-2sin(θ) + sin^{2}(θ)) / ( 1-sin^{2}(θ))

(1-sinθ)/(1+sinθ) = (1-2sin(θ)+sin^{2}(θ)) / (cos^{2}(θ))

(1-sinθ)/(1+sinθ) = 1/cos^{2}(θ) - 2*sin(θ)/cos^{2}(θ) +sin^{2}(θ)/cos^{2}(θ)

(1-sinθ)/(1+sinθ) = (1/cos^{2}(θ)) - 2*tan(θ)/cos(θ) + tan^{2}(θ)

Now expanding using FOIL ((1/cosθ)-tanθ)^2 = ((1/cos(θ))^{2} -2*tan(θ)/cos(θ)+tan^{2}(θ))

((1/cosθ)-tanθ)^2 = (1-sinθ)/(1+sinθ)

Let me know if you need any other help!