Question

. (iii) cos x + sin x = cos 2x + sin 2x​

Answer 1

Cosx+sinx=cos2x+sin2x, then

cosx-cos2x=sin2x-sinx

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

sin(3x/2) sin(x/2) - cos(3x/2) sin(x/2) = 0

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

Either sin(x/2) = 0, or sin(3x/2) - cos(3x/2)=0

So if sin(x/2) = 0, then

sin (x/2) = sin 0

Thus x/2 = nπ +(-1)ⁿ ×0

or x/2 = nπ or x = 2nπ.

And if sin(3x/2) - cos(3x/2) = 0

Dividing both sides by cos(3x/2), we get

tan(3x/2) - 1 = 0

tan(3x/2) = 1

tan(3x/2) = tan(π/4)

3x/2 = nπ + π/4

3x = 2nπ + π/2

x = (2nπ/3) + π/6.

Answer 2

Answer:cosx+sinx=cos2x+sin2x, then

cosx-cos2x=sin2x-sinx

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

sin(3x/2) sin(x/2) - cos(3x/2) sin(x/2) = 0

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

Either sin(x/2) = 0, or sin(3x/2) - cos(3x/2)=0

So if sin(x/2) = 0, then

sin (x/2) = sin 0

Thus x/2 = nπ +(-1)ⁿ ×0

or x/2 = nπ or x = 2nπ.

And if sin(3x/2) - cos(3x/2) = 0

Dividing both sides by cos(3x/2), we get

tan(3x/2) - 1 = 0

tan(3x/2) = 1

tan(3x/2) = tan(π/4)

3x/2 = nπ + π/4

3x = 2nπ + π/2

x = (2nπ/3) + π/6.

Step-by-step explanation: