Question

How do you solve `abs(sinx)=sqrt3/2` in the interval [0,360]?

Answer 1

Solutions: In `[0,360] x= {60^0 ,120^0 ,240^0 ,300^0}`

Explanation:

`|sinx| = sqrt3/2 or sinx = sqrt3/2 ; sin 60^0=sqrt3/2 , sin 120=sqrt 3/2 :. x=60^0 , x=120^0` OR

`|sinx| = sqrt3/2 or sinx = - sqrt3/2 ; sin 240^0= -sqrt3/2 , sin 300= -sqrt 3/2 :. x=240^0 , x=300^0`

Solutions: In `[0,360]x=60^0 , x=120^0 ,x=240^0 , x=300^0` [Ans]

Answer 2

`pi/3, (2pi)/3, (4pi)/3, (5pi)/3`

Explanation:

Separate solving in 2 cases:
a. `sin x = sqrt3/2`
Trig table and unit circle give 2 solution arcs:
`x = pi/3 and x = 2pi/3`
b. `sin x = - sqrt3/2`
Trig table and unit circle give:
`x = - pi/3, and x = - (2pi)/3`
The co-terminal arc of `(-pi/3) is (5pi)/3`
The co-terminal arc of `((-2pi)/3) is (4pi)/3`
Answers for `(0, 2pi)`;
`pi/3, (2pi)/3, (4pi)/3, (5pi)/3`