Cos^2x-sin2x=1 Cos-2sin2x=0 Sin^22x=cos^2x+cos3x 26/09/2021 Bởi Hailey Cos^2x-sin2x=1 Cos-2sin2x=0 Sin^22x=cos^2x+cos3x
1) ${\cos}^2x-\sin2x=1$ $\Rightarrow -\sin2x=1-{\cos}^2x$ $\Rightarrow -\sin2x={\sin}^2x$ $\Rightarrow {\sin}^2x+2\sin x\cos x=0$ $\Rightarrow \sin x(\sin x+2\cos x)=0$ $\Rightarrow \left[ \begin{array}{l} \sin x=0 \\ \sin x+2\cos x=0 \end{array} \right .$ $\Rightarrow \left[ \begin{array}{l} x\ne k\pi (k\in\mathbb Z)\\ \dfrac{1}{\sqrt5}\sin x+\dfrac{2}{\sqrt5}\cos x=0 (1)\end{array} \right .$ Giải $(1)$ đặt $\cos\alpha=\dfrac{1}{\sqrt5}$, $\sin\alpha=\dfrac{2}{\sqrt5}$ $\Rightarrow\cos\alpha\sin x+\sin\alpha\cos x=0 $ $\Rightarrow \sin(x+\alpha)=0$ $\Rightarrow x+\alpha=k\pi\Rightarrow x=-\alpha+k\pi(k\in\mathbb Z)$ 2) $\cos-2\sin2x=0$ đề thiếu. 3) ${\sin}^22x={\cos}^2x+\cos3x$ $1-{\cos}^22x={\cos}^2x+4{\cos}^3x-3\cos x$ $\Rightarrow 1-(2{\cos}^2x-1)^2={\cos}^2x+4{\cos}^3x-3\cos x$ Đặt $\cos x=t$ $\Rightarrow 1-(2t-1)^2+t^2+4t^3-3t$ $\Rightarrow4t^4+4t^3-3t^2-3t=0$ Bình luận
1) ${\cos}^2x-\sin2x=1$
$\Rightarrow -\sin2x=1-{\cos}^2x$
$\Rightarrow -\sin2x={\sin}^2x$
$\Rightarrow {\sin}^2x+2\sin x\cos x=0$
$\Rightarrow \sin x(\sin x+2\cos x)=0$
$\Rightarrow \left[ \begin{array}{l} \sin x=0 \\ \sin x+2\cos x=0 \end{array} \right .$
$\Rightarrow \left[ \begin{array}{l} x\ne k\pi (k\in\mathbb Z)\\ \dfrac{1}{\sqrt5}\sin x+\dfrac{2}{\sqrt5}\cos x=0 (1)\end{array} \right .$
Giải $(1)$ đặt $\cos\alpha=\dfrac{1}{\sqrt5}$, $\sin\alpha=\dfrac{2}{\sqrt5}$
$\Rightarrow\cos\alpha\sin x+\sin\alpha\cos x=0 $
$\Rightarrow \sin(x+\alpha)=0$
$\Rightarrow x+\alpha=k\pi\Rightarrow x=-\alpha+k\pi(k\in\mathbb Z)$
2) $\cos-2\sin2x=0$ đề thiếu.
3) ${\sin}^22x={\cos}^2x+\cos3x$
$1-{\cos}^22x={\cos}^2x+4{\cos}^3x-3\cos x$
$\Rightarrow 1-(2{\cos}^2x-1)^2={\cos}^2x+4{\cos}^3x-3\cos x$
Đặt $\cos x=t$
$\Rightarrow 1-(2t-1)^2+t^2+4t^3-3t$
$\Rightarrow4t^4+4t^3-3t^2-3t=0$