How do you solve the following equation sin 2x - sin x = 0sin2xsinx=0 in the interval [0, 2pi]?

2 Answers
Jun 20, 2016

Use the identity sin2theta = 2sinthetacosthetasin2θ=2sinθcosθ.

Explanation:

2sinxcosx - sinx = 02sinxcosxsinx=0

sinx(2cosx - 1) = 0sinx(2cosx1)=0

sinx = 0 and cosx = 1/2sinx=0andcosx=12

0, pi, pi/3 and (5pi)/30,π,π3and5π3

Hopefully this helps!

Jun 20, 2016

The soln. set ={0,pi/3,pi,5pi/3,2pi}.={0,π3,π,5π3,2π}.

Explanation:

Given that, sin2x-sinx=0. rArr 2sinxcosx-sinx=0. rArr sinx(2cosx-1)=0. rArr sinx=0, or, cosx=1/2sin2xsinx=0.2sinxcosxsinx=0.sinx(2cosx1)=0.sinx=0,or,cosx=12

The zeros of sin are, kpi, k in ZZ. Since we require zeros in [0,2pi], these are 0,pi,2pi.

Next cosx=1/2=cos(pi/3) rArr x=2kpi+-pi/3, k in ZZ.This is since that the general soln. of the eqn. costheta=cosalpha is theta=2kpi+-alpha, k in ZZ.

k=0, rArr x=+pi/3 in [0,2pi], as -pi/3!in[0,2pi]
k=1 rArr x=2pi-pi/3=5pi/3, as 2pi+pi/3 !in [0,2pi].

Altogether, the soln. set ={0,pi/3,pi,5pi/3,2pi}.