Jul 29, 2013 Thread starter Admin #1 anemone MHB POTW Director Staff member Feb 14, 2012 3,967 Solve \(\displaystyle 2\sin^4 (x)(\sin((2x)-3)-2\sin^2 (x)(\sin((2x)-3)-1=0\).
Jul 30, 2013 Admin #2 M MarkFL Administrator Staff member Feb 24, 2012 13,775 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: \(\displaystyle x=\frac{\pi}{2}(2k+1)+\frac{1}{2}\sin^{-1}\left(\sqrt{3}-1 \right)\)
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: \(\displaystyle x=\frac{\pi}{2}(2k+1)+\frac{1}{2}\sin^{-1}\left(\sqrt{3}-1 \right)\)