How do you solve the equation on the interval [0,2pi) for #sin 3x = -sinx#?

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Answer 1 · 2016-08-24T06:18:22

The Soln. Set is #{0,pi/2,pi,3pi/2}#.

We recall that, #sin3x=3sinx-4sin^3x#, so, the given eqn. becomes,

#3sinx-4sin^3x+sinx=0#.

#:. 4sinx-4sin^3x=0#.

#:. 4sinx(1-sin^2x)=0#.

#:. 4sinxcos^2x=0#.

#:. sinx=0, cosx=0#.

#:. x=kpi, or, x=(2k+1)pi/2, k in ZZ#.

Hence, the Soln. Set is #{kpi}uu{(2k+1)pi/2}, k in ZZ#.

#:. "But, x" in [0,2pi), &, sinx=0 rArr x=0,pi#, while,

for #cosx=0, "such x are" pi/2, 3pi/2#.