3cosx -sinx=3 5cos²x-3cosx-2=0 Cos²x-sin²xcosx=0

By Isabelle

3cosx -sinx=3
5cos²x-3cosx-2=0
Cos²x-sin²xcosx=0

0 bình luận về “3cosx -sinx=3 5cos²x-3cosx-2=0 Cos²x-sin²xcosx=0”

  1. 1) $3\cos x -\sin x=3$
    $\Rightarrow \dfrac{3}{\sqrt{10}}\cos x-\dfrac{1}{\sqrt{10}}\sin x=\dfrac{3}{\sqrt{10}}$
    Đặt $\sin\alpha=\dfrac{3}{\sqrt{10}}$ và $\cos\alpha=\dfrac{1}{\sqrt{10}}$
    Phương trình tương đương: $\sin\alpha\cos x-\cos\alpha\sin x=\dfrac{3}{\sqrt{10}}$
    $\Rightarrow \sin(\alpha-x)=\dfrac{3}{\sqrt{10}}$
    $\Rightarrow \left[\begin{array}{l} \alpha-x=\arcsin\dfrac{3}{\sqrt{10}}+k2\pi\\ \alpha-x=\pi-\arcsin\dfrac{3}{\sqrt{10}}+k2\pi \end{array} \right .$
    $\Rightarrow \left[\begin{array}{l} x=\alpha-\arcsin\dfrac{3}{\sqrt{10}}+k2\pi\\ x=\alpha-\pi+\arcsin\dfrac{3}{\sqrt{10}}+k2\pi \end{array} \right .$

    2) $5{\cos}^2x-3\cos x-2=0$
    $\Rightarrow \left[ \begin{array}{l} \cos x=-1 \\ \cos x=\dfrac{2}{5} \end{array} \right .$
    $\Rightarrow \left[ \begin{array}{l} x=\pi+k2\pi \\x=\pm\arccos\dfrac{2}{5}+k2\pi \end{array} \right .(k\in\mathbb Z)$

    3) ${\cos}^2x-{\sin}^2x\cos x=0$
    $\Rightarrow \cos x(\cos x-{\sin}^2x)$
    $\Rightarrow \left\{ \begin{array}{l} \cos x=0\Rightarrow x=\dfrac{\pi}{2}+k\pi(k\in\mathbb Z)\\ \cos x-{\sin}^2x=0(1)\end{array} \right .$
    $(1)\Rightarrow \cos x-(1-{\cos}^2x)=0$
    $\Rightarrow \left[\begin{array}{l} \cos x=\dfrac{-1-\sqrt5}{2} <-1(l)\\ \cos x=\dfrac{-1-\sqrt5}{2} \Rightarrow x=\pm\arccos\dfrac{-1-\sqrt5}{2}+k2\pi(k\in\mathbb Z)\end{array} \right .$

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