Question

I. Find the solutions of the following trigonometric conditional equations for 0 ≤ ϴ ≤ 2π. (Rx2)

1. 4 cos ϴ + 4 = 0
2. sec 2 ϴ = 1
3. sin ϴ - 1 = 0
4. 2 cos ϴ - 1 = 0
5. 4 cos2 ϴ = 3
6. cos ϴ + sin 2ϴ = 0
7. cot 2ϴ = 3
8. cos2 ϴ - cos ϴ = 0
9. sin2 ϴ = ¾
10. csc2 ϴ - 1 = 0

Answer 1

I don't know how to solve the parts with sec 2 ϴ = 1,sec 2 ϴ = 1, csc2 ϴ - 1 = 0 I don't know if it is a typo or it is the real equation.

Answer 2

is it sec^2 or sec(2x) ?

Answer 3

What do you think?

Answer 4

is it possible to solve if it is sec(2x) ????

Answer 5

because I think it is typo

Answer 6

yes
sec(2x) = 1
1/cos(2x) =1
cos(2x) = 1
cos is 1 when angle is 0
2x = 0
x = 0

Answer 7

thanks! I'll just try both

Answer 8

if its squared, then it will be
sec(x) = +-1

Answer 9

w8 I think your solution is incomplete

Answer 10

oh, you are right

2x =2pi
x = pi

Answer 11

cos^2 + sin^2 = 1

Answer 12

isn't it that sec 2A will be 1/(cos^2 A - sin^2 A) ??????????

Answer 13

My teacher taught us that cos2A = cos^2 A - sin^2 A

Answer 14

sorry,
cos^2 - sin^2 = 1
cos^2 = 1+ sin^2
cos^2 = 1+(1-cos^2)
2cos^2 = 2
cos^2 = 1
cos A = +-1

so you end up with same solutions
A = 0,pi

Answer 15

ohhh, thanks!

Answer 16

sure, the csc one will work the same way

Answer 17

another thing is 1/x = 1 also the same as x=1 ??? because when you cross multiply it, it will be the result?

Answer 18

correct