1+sin^3x+cos^3x =3/2sin2x

Đáp án:

$x\in\left\{{-\dfrac{\pi}{2}+k2\pi, \pi +k2\pi}\right\},(k\in\mathbb Z)$

Giải thích các bước giải:

Ta có:

$1+\sin^3x+\cos^3x=\dfrac32\cdot \sin2x$

$\to 1+\sin^3x+\cos^3x=\dfrac32\cdot 2\sin x\cos x$

$\to 1+\sin^3x+\cos^3x=3\sin x\cos x$

$\to 1+(\sin x+\cos x)(\sin^2x-\sin x\cos x+\cos^2x)=3\sin x\cos x$

$\to 1+(\sin x+\cos x)(\sin^2x+\cos^2x-\sin x\cos x)=3\sin x\cos x$

$\to 1+(\sin x+\cos x)(1-\sin x\cos x)=3\sin x\cos x$

Mà $\sin^2x+\cos^2x=1\to (\sin x+\cos x)^2=1+2\sin x\cos x$

Đặt $\sin x+\cos x=a, \sin x\cos x=b$

Ta có $a^2=(\sin x+\cos x)^2\le 2(\sin^2x+\cos^2x)=2$

$\to -\sqrt2\le a\le \sqrt2$

$\to\begin{cases}1+a(1-b)=3b\\ a^2=1+2b\end{cases}$

$\to\begin{cases}1+a(1-\dfrac12(a^2-1))=3\cdot \dfrac12(a^2-1)\\ b=\dfrac12(a^2-1)\end{cases}$

Ta có:

$1+a(1-\dfrac12(a^2-1))=3\cdot \dfrac12(a^2-1)$

$\to \dfrac{3a}{2}-\dfrac{a^3}{2}+1=\dfrac{3}{2}a^2-\dfrac{3}{2}$

$\to -\dfrac{a^3}{2}-\dfrac{3a^2}{2}+\dfrac{3a}{2}+\dfrac{3}{2}+1=0$

$\to -a^3-3a^2+3a+5=0$

$\to -\left(a+1\right)\left(a^2+2a-5\right)=0$

$\to a=-1,\:a=-1+\sqrt{6},\:a=-1-\sqrt{6}$

Mà $-\sqrt2<a<\sqrt 2$

$\to a=-1$

$\to b=0$

$\to \begin{cases}\sin x+\cos x= -1\\ \sin x\cos x=0\end{cases}$

$\to \sin x=-1, \cos x=0\to x=-\dfrac{\pi}{2}+k2\pi,(k\in\mathbb Z)$

Hoặc $\sin x=0, \cos x=-1\to x=\pi +k2\pi,(k\in\mathbb Z)$.