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

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Answer 1 · 2015-05-17T14:57:50

#sin 3x = -1 = sin ((3pi)/2) -> 3x = (3pi)/2 -> x = pi/2.#

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

Sin 3x= -1 would have us 3x= #(3pi)/2# and also #(3pi)/2 +2pi# and #(3pi)/2+4pi#

x= #pi/2#, #pi/2+(2pi)/3#, #pi/2+(4pi)/3#

x=#pi/2, (7pi)/6, (11pi)/6#

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

Answer 2 · 2015-11-10T01:27:38

Solve sin 3x = - 1

Ans: #pi/2; (7pi)/6; (11pi)/6# for #(0, 2pi)#

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

#x = pi/2 + (2/3)kpi#
- If k = 0 --> #x = pi/2#
- If k = 1 --> #x = pi/2 + (2pi)/3 = (7pi)/6#
- If k = 2 --> #x = pi/2 + (4pi)/3 = (11pi)/6#