Câu hỏi của Sonyeondan Bangtan

Question

Giải các pt sau:a) $\cos^2x-\cos x=0$b) $2\sin2x$ + $\sqrt{2}\sin4x=0$ c) $8\cos^2x+2\sin x-7=0$d) $4\cos^4x+\cos^2x-3=0$e) $\sqrt{3}\tan x-6\cot x+\left(2\sqrt{3}-3\right)=0$

Hồng Phúc · Accepted Answer

a, $cos^2x-cosx=0$$\Leftrightarrow cosx\left(cosx-1\right)=0$$\Leftrightarrow\left[{}\begin{matrix}cosx=0\cosx=1\end{matrix}\right.$$\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\x=0\end{matrix}\right.$b, $2sin2x+\sqrt{2}sin4x=0$$\Leftrightarrow2sin2x+2\sqrt{2}sin2x.cos2x=0$$\Leftrightarrow sin2x\left(1+\sqrt{2}cos2x\right)=0$$\Leftrightarrow\left[{}\begin{matrix}sin2x=0\1+\sqrt{2}cos2x=0\end{matrix}\right.$$\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\cos2x=-\dfrac{\sqrt{2}}{2}\end{matrix}\right.$$\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\2x=\dfrac{3\pi}{4}+k2\pi\2x=\dfrac{\pi}{4}+k2\pi\end{matrix}\right.$$\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\x=\dfrac{3\pi}{8}+k\pi\x=\dfrac{\pi}{8}+k\pi\end{matrix}\right.$Câu trả lời: a, $cos^2x-cosx=0$$\Leftrightarrow cosx\left(cosx-1\right)=0$$\Leftrightarrow\left[{}\begin{matrix}cosx=0\cosx=1\end{matrix}\right.$$\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\x=0\end{matrix}\right.$b, $2sin2x+\sqrt{2}sin4x=0$$\Leftrightarrow2sin2x+2\sqrt{2}sin2x.cos2x=0$$\Leftrightarrow sin2x\left(1+\sqrt{2}cos2x\right)=0$$\Leftrightarrow\left[{}\begin{matrix}sin2x=0\1+\sqrt{2}cos2x=0\end{matrix}\right.$$\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\cos2x=-\dfrac{\sqrt{2}}{2}\end{matrix}\right.$$\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\2x=\dfrac{3\pi}{4}+k2\pi\2x=\dfrac{\pi}{4}+k2\pi\end{matrix}\right.$$\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\x=\dfrac{3\pi}{8}+k\pi\x=\dfrac{\pi}{8}+k\pi\end{matrix}\right.$

Lê Ng Hải Anh · Answer

a, $cos^2x-cosx=0$$\Leftrightarrow cosx\left(cosx-1\right)=0$$\Leftrightarrow\left[{}\begin{matrix}cosx=0\cosx=1\end{matrix}\right.$$\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\x=k2\pi\end{matrix}\right.$ (k ∈ Z)Vậy...b, $2sin2x+\sqrt{2}sin4x=0$$\Leftrightarrow2sin2x+2\sqrt{2}sin2x.cos2x=0$$\Leftrightarrow2sin2x\left(1+\sqrt{2}cos2x\right)=0$$\Leftrightarrow\left[{}\begin{matrix}sin2x=0\cos2x=\dfrac{-\sqrt{2}}{2}\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\2x=\pm\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\x=\pm\dfrac{3\pi}{8}+k\pi\end{matrix}\right.$Vậy...c, $8cos^2x+2sinx-7=0$$\Leftrightarrow8\left(1-sin^2x\right)+2sinx-7=0$$\Leftrightarrow8sin^2x-2sinx-1=0$$\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\sinx=-\dfrac{1}{4}\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\x=\dfrac{5\pi}{6}+k2\pi\x=arcsin\left(-\dfrac{1}{4}\right)+k2\pi\x=\pi-arcsin\left(-\dfrac{1}{4}\right)+k2\pi\end{matrix}\right.$Vậy...d, $4cos^4x+cos^2x-3=0$$\Leftrightarrow\left[{}\begin{matrix}cos^2x=\dfrac{3}{4}\cos^2x=-1\left(loai\right)\end{matrix}\right.$ $\Leftrightarrow\dfrac{cos2x+1}{2}=\dfrac{3}{4}$$\Leftrightarrow cos2x=\dfrac{1}{2}$$\Leftrightarrow2x=\pm\dfrac{\pi}{3}+k2\pi$$\Leftrightarrow x=\pm\dfrac{\pi}{6}+k\pi$Vậy...e, $\sqrt{3}tanx-6cotx+\left(2\sqrt{3}-3\right)=0$ (ĐK: $x\ne\dfrac{k\pi}{2}$)$\Leftrightarrow\sqrt{3}tanx-\dfrac{6}{tanx}+\left(2\sqrt{3}-3\right)=0$$\Leftrightarrow\sqrt{3}tan^2x+\left(2\sqrt{3}-3\right)tanx-6=0$$\Leftrightarrow\left[{}\begin{matrix}tanx=\sqrt{3}\tanx=-2\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k\pi\left(tm\right)\x=arctan\left(-2\right)+k\pi\end{matrix}\right.$Vậy... Câu trả lời: a, $cos^2x-cosx=0$$\Leftrightarrow cosx\left(cosx-1\right)=0$$\Leftrightarrow\left[{}\begin{matrix}cosx=0\cosx=1\end{matrix}\right.$$\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\x=k2\pi\end{matrix}\right.$ (k ∈ Z)Vậy...b, $2sin2x+\sqrt{2}sin4x=0$$\Leftrightarrow2sin2x+2\sqrt{2}sin2x.cos2x=0$$\Leftrightarrow2sin2x\left(1+\sqrt{2}cos2x\right)=0$$\Leftrightarrow\left[{}\begin{matrix}sin2x=0\cos2x=\dfrac{-\sqrt{2}}{2}\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\2x=\pm\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\x=\pm\dfrac{3\pi}{8}+k\pi\end{matrix}\right.$Vậy...c, $8cos^2x+2sinx-7=0$$\Leftrightarrow8\left(1-sin^2x\right)+2sinx-7=0$$\Leftrightarrow8sin^2x-2sinx-1=0$$\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\sinx=-\dfrac{1}{4}\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\x=\dfrac{5\pi}{6}+k2\pi\x=arcsin\left(-\dfrac{1}{4}\right)+k2\pi\x=\pi-arcsin\left(-\dfrac{1}{4}\right)+k2\pi\end{matrix}\right.$Vậy...d, $4cos^4x+cos^2x-3=0$$\Leftrightarrow\left[{}\begin{matrix}cos^2x=\dfrac{3}{4}\cos^2x=-1\left(loai\right)\end{matrix}\right.$ $\Leftrightarrow\dfrac{cos2x+1}{2}=\dfrac{3}{4}$$\Leftrightarrow cos2x=\dfrac{1}{2}$$\Leftrightarrow2x=\pm\dfrac{\pi}{3}+k2\pi$$\Leftrightarrow x=\pm\dfrac{\pi}{6}+k\pi$Vậy...e, $\sqrt{3}tanx-6cotx+\left(2\sqrt{3}-3\right)=0$ (ĐK: $x\ne\dfrac{k\pi}{2}$)$\Leftrightarrow\sqrt{3}tanx-\dfrac{6}{tanx}+\left(2\sqrt{3}-3\right)=0$$\Leftrightarrow\sqrt{3}tan^2x+\left(2\sqrt{3}-3\right)tanx-6=0$$\Leftrightarrow\left[{}\begin{matrix}tanx=\sqrt{3}\tanx=-2\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k\pi\left(tm\right)\x=arctan\left(-2\right)+k\pi\end{matrix}\right.$Vậy...

Hồng Phúc · Answer

c, $8cos^2x+2sinx-7=0$$\Leftrightarrow-8sin^2x+2sinx+1=0$$\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\sinx=-\dfrac{1}{4}\end{matrix}\right.$Với $sinx=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.$Với $sinx=-\dfrac{1}{4}\Leftrightarrow\left[{}\begin{matrix}x=arcsin\left(-\dfrac{1}{4}\right)+k2\pi\x=\pi-arcsin\left(-\dfrac{1}{4}\right)+k2\pi\end{matrix}\right.$d, $4cos^4x+cos^2x-3=0$$\Leftrightarrow\left(4cos^2x-3\right)\left(cos^2x+1\right)=0$$\Leftrightarrow4cos^2x-3=0\left(\text{Vì }cos^2x+1>0\right)$$\Leftrightarrow cos^2x=\dfrac{3}{4}$$\Leftrightarrow cosx=\pm\dfrac{\sqrt{3}}{2}$Với $cosx=\dfrac{\sqrt{3}}{2}\Leftrightarrow x=\pm\dfrac{\pi}{3}+k2\pi$Với $cosx=-\dfrac{\sqrt{3}}{2}\Leftrightarrow x=\pm\dfrac{5\pi}{6}+k2\pi$Câu trả lời: c, $8cos^2x+2sinx-7=0$$\Leftrightarrow-8sin^2x+2sinx+1=0$$\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\sinx=-\dfrac{1}{4}\end{matrix}\right.$Với $sinx=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.$Với $sinx=-\dfrac{1}{4}\Leftrightarrow\left[{}\begin{matrix}x=arcsin\left(-\dfrac{1}{4}\right)+k2\pi\x=\pi-arcsin\left(-\dfrac{1}{4}\right)+k2\pi\end{matrix}\right.$d, $4cos^4x+cos^2x-3=0$$\Leftrightarrow\left(4cos^2x-3\right)\left(cos^2x+1\right)=0$$\Leftrightarrow4cos^2x-3=0\left(\text{Vì }cos^2x+1>0\right)$$\Leftrightarrow cos^2x=\dfrac{3}{4}$$\Leftrightarrow cosx=\pm\dfrac{\sqrt{3}}{2}$Với $cosx=\dfrac{\sqrt{3}}{2}\Leftrightarrow x=\pm\dfrac{\pi}{3}+k2\pi$Với $cosx=-\dfrac{\sqrt{3}}{2}\Leftrightarrow x=\pm\dfrac{5\pi}{6}+k2\pi$

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