How do you solve the equation on the interval [0,2pi) for sin 3x = -sinxsin3x=sinx?

1 Answer
Aug 24, 2016

The Soln. Set is {0,pi/2,pi,3pi/2}{0,π2,π,3π2}.

Explanation:

We recall that, sin3x=3sinx-4sin^3xsin3x=3sinx4sin3x, so, the given eqn. becomes,

3sinx-4sin^3x+sinx=03sinx4sin3x+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.