Register Now

Login

Lost Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

222-9+11+12:2*14+14 = ? ( )

Toán Lớp 11: (1+căn 5)(sinx-cosx)+sin2x-1-căn 5=0 thì sinx bằng bao nhiêu

Toán Lớp 11: (1+căn 5)(sinx-cosx)+sin2x-1-căn 5=0 thì sinx bằng bao nhiêu

Comments ( 1 )

  1. Giải đáp: $\sin x\in\{0,1\}$
    Lời giải và giải thích chi tiết:
    Ta có:
    $(1+\sqrt5)(\sin x-\cos x)+\sin2x-1-\sqrt5=0$
    $\to (1+\sqrt5)(\sin x-\cos x)+2\sin x\cos x-(\sin^2x+\cos^2x)-\sqrt5=0$
    $\to (1+\sqrt5)(\sin x-\cos x)-(\sin^2x-2\sin x\cos x+\cos^2x)-\sqrt5=0$
    $\to (1+\sqrt5)(\sin x-\cos x)-(\sin x-\cos x)^2-\sqrt5=0$
    Đặt $\sin x-\cos x=t$
    $\to (1+\sqrt5)t-t^2-\sqrt5=0$
    $\to t\in\{1,\sqrt5\}$
    Mà $t=\sin x-\cos x$
    $\to t\cdot \dfrac1{\sqrt2}=\sin(x-\dfrac14\pi)\le 1$
    $\to t\le\sqrt2$
    $\to t=1$
    $\to \sin x-\cos x=1$
    $\to \sin x-1=\cos x$
    $\to (\sin x-1)^2=\cos^2x$
    $\to\sin^2x-2\sin x+1=1-\sin^2x$
    $\to 2\sin^2x-2\sin x=0$
    $\to 2\sin x(\sin x-1)=0$
    $\to \sin x\in\{0,1\}$

Leave a reply

222-9+11+12:2*14+14 = ? ( )