((Sinx+cosx)^2-2sin^2x)/1+cot^2x=(căn 2)/2 ×[sin(pi/4 -x)-sin(pi/4-3x)]

((Sinx+cosx)^2-2sin^2x)/1+cot^2x=(căn 2)/2 ×[sin(pi/4 -x)-sin(pi/4-3x)]

Share

1 Answer

  1. ĐKXD: $\sin x \neq 0$ hay $x \neq k\pi$.

    Ptrinh tương đương vs

    $\dfrac{\sin^2x + \cos^2x + 2\sin x \cos x – 2\sin^2x}{\dfrac{1}{\sin^2x}} = \dfrac{\sqrt{2}}{2} (\dfrac{\sqrt{2}}{2} (\cos x – \sin x) – \dfrac{\sqrt{2}}{2} (\cos(3x) – \sin(3x)))$

    <->$(1 – 2\sin^2x + \sin(2x)).\sin^2x = \dfrac{1}{2} (\cos x – \sin x – \cos(3x) + \sin(3x))$

    <->$ 2[\sin(2x) + \cos(2x)] . \sin^2x = \cos x – \sin x – \cos(3x) + \sin(3x)$

    <->$ 2[\sin(2x) + \cos(2x)] . \sin^2x = \cos x – \cos(3x) + \sin(3x) – \sin x$

    Áp dụng công thức biến tổng thành tích

    $2[\sin(2x) + \cos(2x)] . \sin^2x = -2 \sin(2x) \sin(-x) + 2\cos(2x) \sin x$

    <->$ [\sin(2x) + \cos(2x)] . \sin^2x = \sin x [\sin(2x) + \cos(2x)]$

    <->$ \sin(2x) + \cos(2x) = 0$ hoặc $\sin^2x = \sin x$

    <->$ \tan(2x) = -1$ hoặc $\sin x = 1$ hoặc $\sin x = 0$ (loại)

    <->$ 2x = -\dfrac{\pi}{4} + k\pi$ hoặc $x = \dfrac{\pi}{2} + 2k\pi$

    <->$ x = -\dfrac{\pi}{8} + k\dfrac{\pi}{2}$ hoặc $x = \dfrac{\pi}{2} + 2k\pi$

    Vậy nghiệm của ptrinh là $ x = -\dfrac{\pi}{8} + k\dfrac{\pi}{2}$ hoặc $x = \dfrac{\pi}{2} + 2k\pi$.

    • 0
Leave an answer

Leave an answer

Browse