How do you solve #sin2xcosx+cos2xsinx= sqrt2/2#?

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Answer 1 · 2015-07-04T05:08:35

#45^@/3# or #pi/12#

We know that

#sinA*cosB+cosA.sinB=Sin(A+B)#

Substitute #A=2x & B=x#

#sinx*cos2x+cosx*sin2x=Sin(x+2x)=sin(3x)##" "color(blue)((1))#

GIven that
#sin2xcosx+cos2xsinx=sqrt2/(2)=(sqrt2)/(sqrt(2)*sqrt(2)#

#sin3x=1/sqrt(2)# { from#" "color(blue)((1))#}

#sin3x=sin(pi/4)#

#cancel(sin)3x=cancel(sin)(pi/4)#

#3x=pi/4#

#x=pi/12#rad

Answer 2 · 2015-07-05T01:11:33

#sin 2xcos x + cos 2xsin x = (sqrt2)/2#

#sin (2x)cos (x) + cos (2x)sin (x) = sin (x + 2x) = sin 3x#

#sin 3x = (sqrt2)/2.#

Trig table gives -> #3x = pi/4 -> x = pi/12#
Trig unit circle gives another arc 3x that has the same sin value

#3x = pi - (pi)/4 = (3pi)/4# -># x = pi/4#

Check with x = pi/4.

#sin (pi/2)cos (pi/4) + cos (pi/2)sin (pi/4) = (sqrt2)/2 + 0 = (sqrt2)/2# . OK