lượng giác

E

eye_smile

Câu 1:

Ta có: $\dfrac{2(1+cot2x.cotx)}{sin^2x}=\dfrac{2}{sin^2x}.\dfrac{sinx.sin2x+cosx.cos2x}{sinx.sin2x}=\dfrac{2}{sin^2x}.\dfrac{cosx}{2sin^2x.cosx}=\dfrac{1}{sin^4x}$

ĐKXĐ:...

PT \Leftrightarrow $\dfrac{1}{sin^4x}+\dfrac{1}{cos^4x}=48$

\Leftrightarrow $sin^4x+cos^4x=48sin^4x.cos^4x$

\Leftrightarrow $1-2sin^2x.cos^2x=48sin^4x.cos^4x$

\Leftrightarrow $sin^2x.cos^2x=\dfrac{1}{8}$

\Leftrightarrow $sin^22x=\dfrac{1}{2}$

\Leftrightarrow ...
 
E

eye_smile

Câu 2:ĐKXĐ:...

Đặt $tan\dfrac{x}{2}=t$

\Rightarrow $sinx=\dfrac{2t}{1+t^2};cosx= \dfrac{1-t^2}{1+t^2}$

Thay vào PT, được :

$6\sqrt{3}t^3+18t^2+14\sqrt{3}t+10=0$

\Leftrightarrow $t= \dfrac{-\sqrt{3}}{3}$

\Leftrightarrow ...
 
D

dien0709

2)Cách khác ĐK:$cosx\neq 0,5$

pt\Leftrightarrow $ 4sinx.sin(x+\dfrac{\pi}{3})+5\sqrt{3}sinx+3(cosx+2)=1-2cosx$

\Leftrightarrow $2sin^2x-1+\sqrt{3}sin2x+5\sqrt{3}sinx+5cosx+6=0$

\Leftrightarrow $-(cos2x-\sqrt{3}sin2x)+5(\sqrt{3}sinx+cosx)+6=0$

\Leftrightarrow $-cos2(x+\dfrac{\pi}{6})+5sin(x+\dfrac{\pi}{6})+3=0$

\Leftrightarrow $2sin^2(x+\dfrac{\pi}{6})+5sin(x+\dfrac{\pi}{6})+2=0$

\Leftrightarrow $sin(x+\dfrac{\pi}{6})=-0,5\to x=-\dfrac{\pi}{3}+k2\pi(l) , x=\pi+k2\pi$

$sin(x+\dfrac{\pi}{6})=-2 (l)$
 
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