cos2x=cosx+sinx find the general solution ?

1 Answer
Aug 31, 2017

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.!