Find all degree solutions in the interval 0° ≤ θ < 360°.

Question

Find all degree solutions in the interval 0° ≤ θ < 360°. If rounding is necessary, round to the nearest tenth of a degree. Use your graphing calculator to verify the solution graphically. (Enter your answers as a comma-separated list.)

10 cos θ + 13 tan θ = sec θ

Answer 1

Hi Shay!

10cos(y) + 13tan(y) = sec(y)

First let's cancel out the quotients. We do this by multiplying both sides by cos(y):

1. cos(y)[10cos(y) + 13tan(y)] = cos(y)[sec(y)] = 1
2. 10cos^2(y) + 13sin(y) = 1
3. 10(1 - sin^2(y)) + 13sin(y) -1 = 0, replacing cos^2(y) with 1 - sin^2(y)√
4. 10 - 10sin^2(y) + 13sin(y) - 1 = 0
5. -10sin^2(y) + 13sin(y) + 9 = 0

Now let sin(y) = x:

1. -10x2 + 13x + 9 = 0

Using the quadratic formula we find that

x = -13 +- √(13^2 - 4(-10)(9)) all divided by 2(-10)

= -13 +- √(169 + 360) all divided by -20

= -13 +-√(529) all divided by -20

= -13 +- 23 all divided by -20

= -10/20 and 36/20

=-1/2 and 9/5

Now since x = sin(y) we find the y values such that sin(y) is either -1/2 or 9/5. sin(y) = -1/2 gives us y = 330 degrees as desired.

I hope this helps!

Rob