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

2 Answers
Binayaka C.
Dec 30, 2016

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]

Nghi N
Dec 30, 2016

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

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