Câu hỏi của Kinder

Question

Giai Pt$\left(2sinx-cosx\right)\left(1+cosx\right)=sin^2x$$3sin^2x+7cos2x-3=0$$\dfrac{4sin^2x+6sin^2x-9-3cos2x}{cosx}=0$

迪丽热巴·迪力木拉提 · Accepted Answer

a/ $\left(2sinx-cosx\right)\left(1+cosx\right)=sin^2x$$\Leftrightarrow\left(2sinx-cosx\right)\left(1+cosx\right)=\dfrac{1-cos2x}{2}$$\Leftrightarrow\left(2sinx-cosx\right)\left(1+cosx\right)=\dfrac{1-2cos^2x+1}{2}=\dfrac{2-2cos^2x}{2}=1-cos^2x$$\Leftrightarrow\left(2sinx-cosx\right)\left(1+cosx\right)=\left(1-cosx\right)\left(1+cosx\right)\Leftrightarrow\left(2sinx-cosx\right)\left(1+cosx\right)-\left(1-cosx\right)\left(1+cosx\right)=0$$\Leftrightarrow\left(1+cosx\right)\left(2sinx-cosx-1+cosx\right)=0\Leftrightarrow\left(1+cosx\right)\left(2sinx-1\right)=0$$\Leftrightarrow\left[{}\begin{matrix}1+cosx=0\2sinx-1=0\end{matrix}\right.$$\Leftrightarrow\left[{}\begin{matrix}cosx=-1\sinx=\dfrac{1}{2}\end{matrix}\right.$$\Leftrightarrow\left[{}\begin{matrix}x=180^o\x=30^o\end{matrix}\right.$  Câu trả lời: a/ $\left(2sinx-cosx\right)\left(1+cosx\right)=sin^2x$$\Leftrightarrow\left(2sinx-cosx\right)\left(1+cosx\right)=\dfrac{1-cos2x}{2}$$\Leftrightarrow\left(2sinx-cosx\right)\left(1+cosx\right)=\dfrac{1-2cos^2x+1}{2}=\dfrac{2-2cos^2x}{2}=1-cos^2x$$\Leftrightarrow\left(2sinx-cosx\right)\left(1+cosx\right)=\left(1-cosx\right)\left(1+cosx\right)\Leftrightarrow\left(2sinx-cosx\right)\left(1+cosx\right)-\left(1-cosx\right)\left(1+cosx\right)=0$$\Leftrightarrow\left(1+cosx\right)\left(2sinx-cosx-1+cosx\right)=0\Leftrightarrow\left(1+cosx\right)\left(2sinx-1\right)=0$$\Leftrightarrow\left[{}\begin{matrix}1+cosx=0\2sinx-1=0\end{matrix}\right.$$\Leftrightarrow\left[{}\begin{matrix}cosx=-1\sinx=\dfrac{1}{2}\end{matrix}\right.$$\Leftrightarrow\left[{}\begin{matrix}x=180^o\x=30^o\end{matrix}\right.$

Lê Thị Thục Hiền · Answer

a) Đáp án: $\left[{}\begin{matrix}cosx=-1\sinx=\dfrac{1}{2}\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}x=\pi+k2\pi\x=\dfrac{\pi}{6}+k2\pi\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.$($k\in Z$)Vậy...b) $3sin^2x+7cos2x-3=0$$\Leftrightarrow3sin^2x+7\left(1-2sin^2x\right)-3=0$$\Leftrightarrow11.sin^2x=4$$\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{2\sqrt{11}}{11}\sinx=\dfrac{-2\sqrt{11}}{11}\end{matrix}\right.$$\Leftrightarrow\left[{}\begin{matrix}x=arc.sin\dfrac{2\sqrt{11}}{11}+k2\pi\x=\pi-arc.sin\dfrac{2\sqrt{11}}{11}+k2\pi\x=arc.sin\dfrac{-2\sqrt{11}}{11}+k2\pi\x=\pi-arc.sin\dfrac{-2\sqrt{11}}{11}+k2\pi\end{matrix}\right.$ ($k\in Z$) (Dị quá,câu này e ko biết đ/a đúng hay sai đâu)Vậy...c)$\dfrac{4.sin^2x+6.sin^2x-9-3.cos2x}{cosx}=0$ (đk: $x\ne\dfrac{\pi}{2}+k\pi$,$k\in Z$)$\Rightarrow10sin^2x-9-3\left(1-2.sin^2x\right)=0$$\Leftrightarrow sin^2x=\dfrac{3}{4}$$\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{\sqrt{3}}{2}\sinx=-\dfrac{\sqrt{3}}{2}\end{matrix}\right.$$\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k2\pi\x=\dfrac{2\pi}{3}+k2\pi\x=\dfrac{-\pi}{3}+k2\pi\x=\dfrac{4\pi}{3}+k2\pi\end{matrix}\right.$($k\in Z$) (Thỏa mãn đk)Vậy...Câu trả lời: a) Đáp án: $\left[{}\begin{matrix}cosx=-1\sinx=\dfrac{1}{2}\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}x=\pi+k2\pi\x=\dfrac{\pi}{6}+k2\pi\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.$($k\in Z$)Vậy...b) $3sin^2x+7cos2x-3=0$$\Leftrightarrow3sin^2x+7\left(1-2sin^2x\right)-3=0$$\Leftrightarrow11.sin^2x=4$$\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{2\sqrt{11}}{11}\sinx=\dfrac{-2\sqrt{11}}{11}\end{matrix}\right.$$\Leftrightarrow\left[{}\begin{matrix}x=arc.sin\dfrac{2\sqrt{11}}{11}+k2\pi\x=\pi-arc.sin\dfrac{2\sqrt{11}}{11}+k2\pi\x=arc.sin\dfrac{-2\sqrt{11}}{11}+k2\pi\x=\pi-arc.sin\dfrac{-2\sqrt{11}}{11}+k2\pi\end{matrix}\right.$ ($k\in Z$) (Dị quá,câu này e ko biết đ/a đúng hay sai đâu)Vậy...c)$\dfrac{4.sin^2x+6.sin^2x-9-3.cos2x}{cosx}=0$ (đk: $x\ne\dfrac{\pi}{2}+k\pi$,$k\in Z$)$\Rightarrow10sin^2x-9-3\left(1-2.sin^2x\right)=0$$\Leftrightarrow sin^2x=\dfrac{3}{4}$$\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{\sqrt{3}}{2}\sinx=-\dfrac{\sqrt{3}}{2}\end{matrix}\right.$$\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k2\pi\x=\dfrac{2\pi}{3}+k2\pi\x=\dfrac{-\pi}{3}+k2\pi\x=\dfrac{4\pi}{3}+k2\pi\end{matrix}\right.$($k\in Z$) (Thỏa mãn đk)Vậy...

迪丽热巴·迪力木拉提 · Answer

b/$3sin^2x+7cos2x-3=0\Leftrightarrow3sin^2x+7\left(2cos^2x-1\right)-3=0\Leftrightarrow3sin^2x+14cos^2x-7-3=0$$\Leftrightarrow3sin^2x+3cos^2x+11cos^2x-10=0\Leftrightarrow3+11cos^2x-10=0\Leftrightarrow11cos^2x-7=0$$\Leftrightarrow cos^2x=\dfrac{7}{11}\Leftrightarrow cosx=\sqrt{\dfrac{7}{11}}$$\Leftrightarrow x=37^o5'$ Ủa sao kết quả xấu vậy:vvv Chắc sai đâu rồi:vvCâu trả lời: b/$3sin^2x+7cos2x-3=0\Leftrightarrow3sin^2x+7\left(2cos^2x-1\right)-3=0\Leftrightarrow3sin^2x+14cos^2x-7-3=0$$\Leftrightarrow3sin^2x+3cos^2x+11cos^2x-10=0\Leftrightarrow3+11cos^2x-10=0\Leftrightarrow11cos^2x-7=0$$\Leftrightarrow cos^2x=\dfrac{7}{11}\Leftrightarrow cosx=\sqrt{\dfrac{7}{11}}$$\Leftrightarrow x=37^o5'$ Ủa sao kết quả xấu vậy:vvv Chắc sai đâu rồi:vv

Ôn tập cuối năm môn Đại số 11

Khoá học trên OLM (olm.vn)

Khoá học trên OLM (olm.vn)