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Calculus math help.?
1) Solve the equation on the interval (0, 2pi). Give exact values (Ex: pi/3).
cos^2 x = cos x
2)Simplify: csc x - tan x sin x cos x
I have no idea how to solve these.
2 Answers
- Roger the MoleLv 75 years agoFavorite Answer
[There's not a bit of calculus in this. It's all trigonometry.]
cos^2 x - cos x = 0
Factor:
cos x (cos x - 1) = 0
Use the Zero Product Principle:
cos x = 0
x = arccos 0 = pi/2 and 3pi/2
cos x - 1 = 0
cos x = 1
x = arccos 1 = 0 and 2pi
So altogether there are solutions at x = 0, pi/2, 3pi/2, and 2pi
2)
csc x - tan x sin x cos x
csc x - (sin x / cos x) sin x cos x
csc x - (sin x) sin x
csc x - sin^2 x
or maybe: (1/sin x) - sin^2 x
- cidyahLv 75 years ago
1)
cos^2 x = cos x
cos x(cos x -1) = 0
cos x = 0
x = pi/2, 3pi/2
cos x - 1 = 0
cos x = 1
You have an open interval (0, 2pi)
There are no solutions for cos x = 1
x= pi/2, 3pi/2
2)
csc x - tan x sin x cos x
= 1 / sin x - (sin x /cos x) sin x cos x
= 1/sin x - sin^2 x
