How do you solve sin(3x)= -1sin(3x)=1 with domain between 0 and 2pi?

2 Answers
May 17, 2015

sin 3x = -1 = sin ((3pi)/2) -> 3x = (3pi)/2 -> x = pi/2.sin3x=1=sin(3π2)3x=3π2x=π2.

Check: When x = pi/2 -> 3x - = (3pi)/2 -> sin 3x = -1x=π23x=3π2sin3x=1 OK

Sin 3x= -1 would have us 3x= (3pi)/23π2 and also (3pi)/2 +2pi3π2+2π and (3pi)/2+4pi3π2+4π

x= pi/2π2, pi/2+(2pi)/3π2+2π3, pi/2+(4pi)/3π2+4π3

x=pi/2, (7pi)/6, (11pi)/6π2,7π6,11π6

These are the three solution for x in 0<=x<=2pi0x2π

Nov 10, 2015

Solve sin 3x = - 1

Ans: pi/2; (7pi)/6; (11pi)/6π2;7π6;11π6 for (0, 2pi)(0,2π)

Explanation:

Trig Table of Special Arcs and trig unit circle -->
sin 3x = - 1 = sin ((3pi)/2)sin3x=1=sin(3π2) --> 3x = (3pi)/2 + 2kpi 3x=3π2+2kπ-->

x = pi/2 + (2/3)kpix=π2+(23)kπ
- If k = 0 --> x = pi/2x=π2
- If k = 1 --> x = pi/2 + (2pi)/3 = (7pi)/6x=π2+2π3=7π6
- If k = 2 --> x = pi/2 + (4pi)/3 = (11pi)/6x=π2+4π3=11π6