cos2x +sin (x+pi /6)=0 căn 2 (sinx +cosx )=tanx +cotx

Question

cos2x +sin (x+pi /6)=0
căn 2 (sinx +cosx )=tanx +cotx

in progress 0
Autumn 4 tháng 2021-07-29T10:54:42+00:00 1 Answers 7 views 0

Answers ( )

    0
    2021-07-29T10:56:10+00:00

    Đáp án:

    1, $x=\dfrac{2\pi}{9}+k2\pi,k\in Z $

    2, $ x=\dfrac{\pi }{4}+k2\pi ,k\in Z$

    Giải thích các bước giải:

    1, $\cos 2x+\sin (x+\dfrac{\pi }{6})=0\Rightarrow \sin(2x+\dfrac{\pi }{2})+\sin (x+\dfrac{\pi }{6})=0$

    $\Rightarrow \sin(2x+\dfrac{\pi }{2})=-\sin (x+\dfrac{\pi }{6})\Rightarrow \sin(2x+\dfrac{\pi }{2})=\sin (-x-\dfrac{\pi }{6})$

    $\Rightarrow 2x+\dfrac{\pi }{2}=-x-\dfrac{\pi }{6}\Rightarrow 3x=\dfrac{2\pi}{3}\Rightarrow x=\dfrac{2\pi}{9}+k2\pi ,k\in Z$

    2, $\sqrt{2}(\sin x+\cos x)=\tan x+\cot x  (\sin 2x\neq 0)$
    $\Rightarrow \sqrt{2}(\sin x+\cos x)=\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x}$
    $\Rightarrow \sqrt{2}(\sin x+\cos x)=\dfrac{1}{\sin x\cos x}$

    Đặt $t=\sin x+\cos x=\sqrt{2}\sin (x+\dfrac{\pi }{4}), \left | t \right | \leq \sqrt{2}$

    Ta được:

    $\sqrt{2}t=\dfrac{1}{\dfrac{t^2-1}{2}}\Leftrightarrow \sqrt{2}t(t^2-1)=2\Leftrightarrow t^3-t-\sqrt{2}=0\Leftrightarrow t=\sqrt{2}$

    $\Rightarrow \sin (x+\dfrac{\pi }{4})=1\Leftrightarrow x=\dfrac{\pi }{4}+k2\pi ,k\in Z$

Leave an answer

Browse

35:5x4+1-9:3 = ? ( )