[toán 11] gpt lượng giác

N

newstarinsky


  • 2)ĐK $sin2x\not=0$
    PT tương đương
    $3[sinx(cos^2x+1)+cosx(sin^2x+1)]=10sinx.cosx\\
    \Leftrightarrow 3[sinx.cosx(sinx+cosx)+sinx+cosx]=10sinx.cosx\\
    \Leftrightarrow 3(sinx.cosx+1)(sinx+cosx)=10sinx.cosx$
    Đặt $t=sinx+cosx\Rightarrow sinx.cosx=\dfrac{t^2-1}{2}$
    PT trở thành
    $3(\dfrac{t^2-1}{2}+1).t=10.\dfrac{t^2-1}{2}\\
    \Leftrightarrow 3t^3-10t^2+3t+10=0\\
    \Leftrightarrow (t-2)(3t^3-4t-5)=0$
    OK nhé
$3)3sin^4x+5cos^4x=3\\
\Leftrightarrow 5cos^4x=3(1-sin^2x)(1+sin^2x)=3cos^2x(1+sin^2x)\\
\Leftrightarrow cos^2x(3+3sin^2x-5cos^2x)=0\\
\Leftrightarrow cos^2x(8sin^2x-2)=0$
 
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