Question

How do you solve `sin(2x)cos(x)=sin(x)`?

Answer 1

`x=npi,2npi+-(pi/4), and 2npi+-((3pi)/4)` where `n in ZZ`

Explanation:

`rarrsin2xcosx=sinx`

`rarr2sinx*cos^2x-sinx=0`

`rarrsinx(2cos^2x-1)=0`

`rarrrarrsinx*(sqrt2cosx+1)*(sqrt2cosx-1)=0`

When `sinx=0`

`rarrx=npi`

When `sqrt2cosx+1=0`

`rarrcosx=-1/sqrt2=cos((3pi)/4)`

`rarrx=2npi+-((3pi)/4)`

When `sqrt2cosx-1=0`

`rarrcosx=1/sqrt2=cos(pi/4)`

`rarrx=2npi+-(pi/4)`

Answer 2

`x = npi, pi/4 + npi, (3pi)/4 + npi` where `n in ZZ`

Explanation:

We have,

`color(white)(xxx)sin2xcosx = sinx`

`rArr 2sinxcosx xx cosx = sinx` [As, `sin 2x = 2sinxcosx`]

`rArr 2sinxcos^2x - sin x = 0`

`rArr sinx(2cos^2 - 1) = 0`

Now,

Either,

`sin x = 0 rArr x = sin^-1(0) = npi`, where `n in ZZ`

Or,

`color(white)(xxx)2cos^2x - 1 = 0`

`rArr 2cos^2x - (sin^2x + cos^2x) = 0` [As `sin^2x + cos^2 x = 1`]

`rArr 2cos^2x-sin^2x-cos^2x = 0`

`rArr cos^2x - sin^2x = 0`

`rArr (cosx + sin x)(cos x - sin x) = 0`

So, Either `cos x - sin x = 0 rArr cos x = sin x rArr x = pi/4 +- npi`, where `n in ZZ`

Or,

`cos x + sin x = 0 rArr cos x = -sinx rArr x = (3pi)/4 +- npi`, where `n in ZZ`

So, Summing it all up,

`x = npi, pi/4 +- npi, (3pi)/4 +- npi`, where `n in ZZ`