Answer:
a) {0.63450, 5.63968}
b) {0.52360 (=π/6), 2.61799 (=5π/6), 3.34395, 6.08183}
e) {1.23096, 2.09440 (=2π/3), 4.18879 (=4π/3), 5.05223}
Explanation:
You want solutions to three quadratic equations involving trig functions.
Identities
These identities are useful in the solution process.
cos²(x) = 1 -sin²(x)
cos(2x) = 2cos²(x) -1
a) 5cos²(x) +6cos(x) = 8
Using z = cos(x) and putting this in standard form we have ...
5z² +6z -8 = 0
(5z +10)(5z -4)/5 = 0 . . . . factor
(z +2)(5z -4) = 0 . . . . simplify
z = -2, z = 4/5 . . . . . . -2 is extraneous, as |cos(x)| ≤ 1
x = {arccos(4/5), 2π-arccos(4/5)} ≈ {0.63450, 5.63968}
b) -10cos²(x) -3sin(x) = -9
Using z = sin(x) and the cos² identity, we have ...
-10(1 -z²) -3z +9 = 0
10z² -3z -1 = 0 . . . . . . . . . . . simplify
(10z -5)(10z +2)/10 = 0 . . . . factor
(2z -1)(5z +1) = 0 . . . . . . . . simplify
z = 1/2 or -1/5
x = arcsin(1/2) or arcsin(-1/5) and π less those values
x ≈ {π/6, 5π/6, 3.34295, 6.08183}
e) 3cos(2x) +2 +cos(x) = 0
Using z = cos(x) and the double-angle identity, we have ...
3(2z² -1) +2 +z = 0
6z² +z -1 = 0 . . . . . . . . . . . simplify
(6z +3)(6z -2)/6 = 0 . . . . . factor
(2z +1)(3z -1) = 0 . . . . . . . simplify
z = -1/2 or 1/3
x = arccos(-1/2) or arccos(1/3) and 2π less these values
x ≈ {1.23096, 2.09440, 4.18879, 5.05223}
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Additional comment
The first two attachments show the calculator results. (Some of the numbers may not be in the desired interval.)
The remaining three attachments show graphing calculator solutions. The graphing calculator provides a quick and easy solution directly in radians, though maybe not to very many decimal places. (The solutions can be refined to as much numerical precision as you like.)
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