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 $x$ .

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