Giải phương trình sau: sin^4x + cos^4x + sin2x -3/2 =0 26/09/2021 Bởi Adeline Giải phương trình sau: sin^4x + cos^4x + sin2x -3/2 =0
${\sin }^4x+{\cos}^4x+\sin 2x-\dfrac{3}{2}=0$ $\Rightarrow ({\sin}^2x)^2+({\cos}^2x)^2+2{\sin}^2x{\cos}^2x-2{\sin}^2x{\cos}^2x+\sin2x-\dfrac{3}{2}=0$ $\Rightarrow ({\sin}x^2+{\cos}^2x)^2-2{\sin}^2x{\cos}^2x+\sin2x-\dfrac{3}{2}=0$ $\Rightarrow 1-2{\sin}^2x{\cos}^2x+\sin2x-\dfrac{3}{2}=0$ $\Rightarrow -\dfrac{1}{2}-2(\dfrac{\sin2x}{2})^2+\sin2x=0$ $\Rightarrow -\dfrac{1}{2}{\sin}^22x+\sin2x-\dfrac{1}{2}=0$ $\Rightarrow \sin2x=1$ $\Rightarrow 2x=\dfrac{\pi}{2}+k2\pi$ $\Rightarrow x=\dfrac{\pi}{4}+k\pi$ ($k\in\mathbb Z$). Bình luận
${\sin }^4x+{\cos}^4x+\sin 2x-\dfrac{3}{2}=0$
$\Rightarrow ({\sin}^2x)^2+({\cos}^2x)^2+2{\sin}^2x{\cos}^2x-2{\sin}^2x{\cos}^2x+\sin2x-\dfrac{3}{2}=0$
$\Rightarrow ({\sin}x^2+{\cos}^2x)^2-2{\sin}^2x{\cos}^2x+\sin2x-\dfrac{3}{2}=0$
$\Rightarrow 1-2{\sin}^2x{\cos}^2x+\sin2x-\dfrac{3}{2}=0$
$\Rightarrow -\dfrac{1}{2}-2(\dfrac{\sin2x}{2})^2+\sin2x=0$
$\Rightarrow -\dfrac{1}{2}{\sin}^22x+\sin2x-\dfrac{1}{2}=0$
$\Rightarrow \sin2x=1$
$\Rightarrow 2x=\dfrac{\pi}{2}+k2\pi$
$\Rightarrow x=\dfrac{\pi}{4}+k\pi$ ($k\in\mathbb Z$).