Solve for 0 < Θ < 2pi
1. 2secΘ + 9 = 5cosΘ
Solve for the same, but without using a calculator.
1. root3tan^2Θ + tanΘ = 0
please help? thanks
Call t = theta for convenience.
sec(t) = 1/cos(t), so
2/cos(t) + 9 = 5cos(t) <-- Multiply both sides by cos(t):
2 + 9cos(t) = 5cos^2(t)
5cos^2(t) - 9cos(t) - 2 = 0
Let x = cos(t)
5x^2 - 9x - 2 = 0
(5x + 1)(x - 2) = 0
x = -1/5 or x = 2
Thus
cos(t) = -1/5 or cos(t) = 2 <-- Impossible
Thus
cos(t) = -1/5
Thus
t = acs(-1/5) = 1.772 rad = 0.5641(pi)
Now, this angle is slightly larger than (pi)/2, so it is in quadrant II. The reference angle is (pi) - 0.5641(pi) = 0.4359(pi). There is another such angle where cosine is negative: Quadrant III. So the other angle will be (pi) + 0.4359(pi) = 1.4360(pi). Thus your answers are:
t = 0.5641(pi) or t = 1.4360(pi)
-Dan
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