#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.!