Find all the solutions in the interval [0,2pi) sin 3x-sin x=0 ?

1 Answer
Jim G.
Feb 5, 2018

#x=pi/4,(3pi)/4,pi,(5pi)/4,(7pi)/4#

Explanation:

#"using the "color(blue)"sum to product formula"#

#•color(white)(x)sinC-sinD=2cos((C+D)/2)sin((C-D)/2)#

#"here "A=3x" and "B=x#

#rArrsin3x-sinx=2cos2xsinx=0#

#rArrcos2x=0" or "sinx=0#

#color(blue)"solving "sinx=0#

#rArrx=cancel(0),pi,cancel(2pi)#

#color(blue)"solving "cos2x=0#

#rArr2x=pi/2,(3pi)/2,(5pi)/2,(7pi)/2#

#rArrx=pi/4,(3pi)/4,(5pi)/4,(7pi)/4#

#"putting the 2 solutions together gives"#

#x=pi/4,(3pi)/4,pi,(5pi)/4,(7pi)/4to(0,2pi)#

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