Môn Toán Lớp 11 Giải phương trình sau 1+3sin2x=2tanx

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Môn Toán Lớp 11 Giải phương trình sau
1+3sin2x=2tanx Giúp em bài này với ạ em cần gấp, đừng copy nguồn trên mạng nha. Em xin cảm ơn thầy cô và các bạn nhiều.

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Nevaeh 2 tuần 2022-09-17T17:53:33+00:00 2 Answers 0 views 0

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    0
    2022-09-17T17:54:54+00:00

    Đáp án:

    $x\in\{\dfrac34\pi+k\pi,\dfrac{8k\pi-\pi +4\arcsin \left(\dfrac{\sqrt{2}}{6}\right)}{8}, \dfrac{3\pi +8k\pi-4\arcsin \left(\dfrac{\sqrt{2}}{6}\right)}{8}\}$

    Giải thích các bước giải:

    ĐKXĐ: $x\ne \dfrac{\pi}{2}+k\pi, k\in Z$

    Ta có:

    $1+3\sin2x=2\tan x$

    $\to 3+3\sin2x=2\tan x+2$

    $\to 3\left(1+\sin2x\right)=2\left(\tan x+1\right)$

    $\to 3\left(\sin^2x+\cos^2x+2\sin x\cos x\right)=2\left(\dfrac{\sin x}{\cos x}+1\right)$

    $\to 3\left(\sin x+\cos x\right)^2=2\cdot\dfrac{\sin x+\cos x}{\cos x}$

    $\to 3\left(\sin x+\cos x\right)^2-2\cdot\dfrac{\sin x+\cos x}{\cos x}=0$

    $\to \left(\sin x+\cos x\right)\left(3\left(\sin x+\cos x\right)-\dfrac{2}{\cos x}\right)=0$

    $\to \left(\sin x+\cos x\right)\left(3\left(\sin x\cos x+\cos^2x\right)-2\right)=0$

    $\to \left(\sin x+\cos x\right)\left(\dfrac32\left(2\sin x\cos x+2\cos^2x-1+1\right)-2\right)=0$

    $\to \left(\sin x+\cos x\right)\left(\dfrac32\left(\sin2x+\cos2x+1\right)-2\right)=0$

    $\to \left(\sin x+\cos x\right)\left(3\left(\sin2x+\cos2x+1\right)-4\right)=0$

    $\to \left(\sin x+\cos x\right)\left(3\left(\sin2x+\cos2x\right)-1\right)=0$

    $\to \sin x+\cos x=0$

    $\to \sin x=-\cos x$

    $\to \dfrac{\sin x}{\cos x}=-1$

    $\to \tan x=-1$

    $\to x=\dfrac34\pi+k\pi$

    Hoặc $3\left(\sin2x+\cos2x\right)-1=0$

    $\to \sin2x+\cos2x=\dfrac13$

    $\to \sin2x\cdot\dfrac{\sqrt{2}}{2}+\cos2x\cdot\dfrac{\sqrt{2}}{2}=\dfrac{\sqrt{2}}{6}$

    $\to \sin2x\cdot\cos\left(\dfrac{\pi}{4}\right)+\cos2x\cdot\sin\left(\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{6}$

    $\to \sin\left(2x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{6}$

    $\to 2x+\dfrac{\pi}{4}=\arcsin\dfrac{\sqrt{2}}{6}+k2\pi\to x=\dfrac{8k\pi-\pi +4\arcsin \left(\dfrac{\sqrt{2}}{6}\right)}{8}$

    Hoặc $2x+\dfrac{\pi}{4}=\pi-\arcsin\dfrac{\sqrt{2}}{6}+k2\pi\to x=\dfrac{3\pi +8k\pi-4\arcsin \left(\dfrac{\sqrt{2}}{6}\right)}{8}$

    0
    2022-09-17T17:55:00+00:00

    Đáp án:

    $\left\{\begin{array}{I}x=-\dfrac{\pi}4+k\pi\\x=\arctan\dfrac{3+\sqrt{17}}4+k\pi\end{array}\right.$ $(k\in\mathbb Z)$

    Giải thích các bước giải:

    $1+3\sin2x=2\tan x$

    Điều kiện $\cos x\ne0\Leftrightarrow x\ne\dfrac{\pi}2+k\pi$ $(k\in\mathbb Z)$

    Phương trình tương đương:

    $1+6\sin x\cos x=2\tan x$

    $\Leftrightarrow\dfrac1{\cos^2x}+6\tan x=2\tan x.dfrac1{\cos^2x}$

    $\Leftrightarrow(1+\tan^2x)+6\tan x=2\tan x(1+\tan^2x)$

    $\Leftrightarrow2\tan^3x-\tan^2x-4\tan x-1=0$

    $\Leftrightarrow(\tan x+1)(2\tan^2x-3\tan x-1)=0$

    $\Leftrightarrow\left[\begin{array}{I}\tan x=-1\\\tan x=\dfrac{3\pm\sqrt{17}}4\end{array}\right.$

    $\Leftrightarrow\left[\begin{array}{I}x=-\dfrac{\pi}4+k\pi\\x=\arctan\dfrac{3+\sqrt{17}}4+k\pi\end{array}\right.$ $(k\in\mathbb Z)$

    Vậy $\left\{\begin{array}{I}x=-\dfrac{\pi}4+k\pi\\x=\arctan\dfrac{3+\sqrt{17}}4+k\pi\end{array}\right.$ $(k\in\mathbb Z)$.

    mon-toan-lop-11-giai-phuong-trinh-sau-1-3sin2-2tan

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