Question

Question #af5d8

Answer 1

a. cos 2x = 0
Unit circle gives:
`2x = pi/2 + 2kpi` and `2x = (3pi)/2 + 2kpi`

`2x = pi/2 + 2kpi` --> `x = pi/4 + kpi`
`2x = (3pi)/2 + 2kpi` --> `x = (3pi)/4 + kpi`
Answers for `(0, 2pi)`:
`pi/4; (3pi)/4; (5pi)/4; (7pi)/4`

b. `cos x = cos 2x = 2cos ^2 x - 1`
Solve this quadratic equation for cos x:
`2cos^2 x - cos x - 1 = 0`
Since a + b + c = 0, use shortcut.
The 2 real roots are: cos x = 1 and cos x = c/a = - 1/2
Trig table and unit circle give:
1/ cos x = 1 --> x = 0 and x = 2kpi
2/ `cos x = - 1/2` --> `x = +- (2pi)/3 + 2kpi`
Answers for `(0, 2pi)`:
`2pi; +- (2pi)/3`
c. sin 3x = 0
Unit circle gives as solutions:
1/ `3x = 0 + 2kpi` --> `x = (2kpi)/3`
2/ `3x = pi + 2kpi`--> `x = ((2k + 1)pi)/3`
3/ 3x = 2pi + 2kpi --> x = (2kpi)/3
Answers for `(0, 2pi)`:
`(2pi)/3; (4pi)/3; 2pi; pi/3; pi`