How do you solve #\sin 2x \cos x + \cos ^ { 2} 2x \sin x = - 1#?

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Answer 1 · 2017-08-01T02:00:28

See below.

#sin2x*cosx+(cos2x)^2*sinx=-1#

#2sinx*(cosx)^2+[1-2(sinx)^2]^2*sinx=-1#

#2sinx*[1-(sinx)^2]+[1-2(sinx)^2]^2*sinx=-1#

After using #u=sinx# transform, this equation became;

#2u*(1-u^2)+(1-2u^2)^2*u=-1#

#2u-2u^3+(4u^4-4u^2+1)*u=-1#

#2u-2u^3+4u^5-4u^3+u=-1#

#4u^5-6u^3+3u+1=0#

#(u+1)*(4u^4-4u^3-2u^2+2u+1)=0#

No real solution from the second multiplier. From first one, #u=-1#.

Due to #sinx=-1#, #x# must be equal to #(3pi)/2+2pi*k#

1) I rewrote this equation in terms of #u=sinx#

2) I solved it or #u#.

3) I found #x#.