Find the general value of x if: cosx+sinx=cos2x+sin2x, using sinc-sind and cosc-cosd formula?

1 Answer
Mar 13, 2018

x in {2kpi}uu{(4k+1)pi/6}, k in ZZ.

Explanation:

cosx+sinx=cos2x+sin2x.

:. cosx-cos2x=sin2x-sinx.

:. -2sin((x+2x)/2)sin((x-2x)/2)=2cos((2x+x)/2)sin((2x-x)/2).

:. +sin(3/2x)sin(+1/2x)=cos(3/2x)sin(1/2x).

:. sin(3/2x)sin(1/2x)-cos(3/2x)sin(1/2x)=0.

:. sin(1/2x)[sin(3/2x)-cos(3/2x)]=0.

:. sin(1/2x)=0, or, sin(3/2x)=cos(3/2x).

Case 1 : sin(1/2x)=0.

sin(1/2x)=0 rArr 1/2x=kpi rArr x=2kpi, k in ZZ.

Case 2 : sin(3/2x)=cos(3/2x).

Note that cos(3/2x)" can not be "0, because, in that case, by the

virtue of the eqn., sin(3/2x)" will also be "0, contradicting,

sin^2(3/2x)+cos^2(3/2x)=1.

So, dividing by cos(3/2x)ne0, we get,

tan(3/2x)=1=tan(pi/4)," giving, "

3/2x=kpi+pi/4 rArr x=2/3kpi+pi/6=(4k+1)pi/6, k in ZZ.

Altogether, The Soln. Set ={2kpi}uu{(4k+1)pi/6}, k in ZZ.