Trigonometric Challenge

[anemone]

Solve \(\displaystyle 2\sin^4 (x)(\sin((2x)-3)-2\sin^2 (x)(\sin((2x)-3)-1=0\).

 

[MarkFL]

My solution:

Spoiler

Factor the first two terms:

\(\displaystyle 2\sin^2(x)\left(2\sin(2x)-3) \right)\left(\sin^2(x)-1 \right)-1=0\)

Apply a Pythagorean identity to the second factor of the first term and multiply through by $-2$:

\(\displaystyle 4\sin^2(x)\cos^2(x)\left(2\sin(2x)-3) \right)+2=0\)

To the first three factors of the first term, apply the double-angle identity for sine:

\(\displaystyle \sin^2(2x)\left(2\sin(2x)-3) \right)+2=0\)

Distribute to obtain a cubic in $\sin(2x)$:

\(\displaystyle 2\sin^3(2x)-3\sin^2(2x)+2=0\)

Factor:

\(\displaystyle \left(\sin(2x)-1 \right)\left(\sin^2(2x)-2\sin(2x)-2 \right)=0\)

Apply the zero factor property:

i) The first factor implies

\(\displaystyle \sin(2x)=1\)

\(\displaystyle x=\frac{\pi}{4}(4k+1)\) where \(\displaystyle k\in\mathbb{Z}\)

ii) The second factor implies, by applying the quadratic formula:

\(\displaystyle \sin(2x)=1\pm\sqrt{3}\)

Discarding the root whose magnitude is greater than unity, we are left with:

\(\displaystyle \sin(2x)=1-\sqrt{3}\)

\(\displaystyle x=k\pi-\frac{1}{2}\sin^{-1}\left(\sqrt{3}-1 \right)\)

Using the identity $\sin(\pi-\theta)=\sin(\theta)$ we also have:

\(\displaystyle x=\frac{\pi}{2}(2k+1)+\frac{1}{2}\sin^{-1}\left(\sqrt{3}-1 \right)\)