Question

`cos2x=cosx+sinx` find the general solution ?

Answer 1

The Soln. Set is, `{npi-pi/4}uu{2npi}uu{2npi-pi/2}, n in ZZ.`

Explanation:

Since, `cos2x=cos^2x-sin^2x=(cosx+sinx)(cosx-sinx),`

we have,

`cos2x=cosx+sinx.`

`rArr (cosx+sinx)(cosx-sinx)-(cosx+sinx)=0.`

`rArr (cos+sinx)(cosx-sinx-1)=0.`

`rArr cos+sinx=0, or, cosx-sinx=1.`

In case, `cosx+sinx=0 :. sinx=-cosx rArr sinx/cosx=-1, (cosxne0).`

`rArr tanx=-1=tan(-pi/4).`

We know that, `tantheta=tanalpha rArr theta=npi+alpha, n in ZZ.`

`:. tanx=tan(-pi/4) rArr x=npi-pi/4, n in ZZ.`

If, `cosx-sinx=1, :. 1/sqrt2(cosx-sinx)=1/sqrt2.`

`:. cosx*1/sqrt2-sinx*1/sqrt2=1/sqrt2.`

`:. cosxcos(pi/4)-sinxsin(pi/4)=1/sqrt2.`

`:. cos(x+pi/4)=1/sqrt2=cos(pi/4).`

We know, `costheta=cosalpharArr theta=2npipmalpha, n in ZZ.`

`:. cos(x+pi/4)=cos(pi/4) rArr x+pi/4=2npipmpi/4.`

`rArr x=2npipmpi/4-pi/4=2npi, or, 2npi-pi/2, n in ZZ.`

Altogether, The Soln. Set is,

`{npi-pi/4}uu{2npi}uu{2npi-pi/2}, n in ZZ.`

Enjoy Maths.!