2 câu trả lời
Đáp án:
$x=\dfrac{\pi}{4}+k\pi$ $(k\in\mathbb Z)$
Giải thích các bước giải:
Ta có:
$\sin^4x+\cos^4x=\sin2x-\dfrac12$
$\Leftrightarrow(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=\sin2x-\dfrac12$
$\Leftrightarrow 1-2.\dfrac{\sin^22x}4=\sin2x-\dfrac12$
$\Leftrightarrow\sin^22x+2\sin2x-3=0$
$\Leftrightarrow\left[\begin{array}{I}\sin 2x=-3<-1\text { (loại)}\\\sin2x=1\Leftrightarrow 2x=\dfrac{\pi}2+k2\pi\end{array}\right.$
$\Leftrightarrow x=\dfrac{\pi}{4}+k\pi$ $(k\in\mathbb Z)$.
sin^4x + cos^4x = sin2x - 1/2
<=> (sin^2x + cos^2x)^2 - 2sin^2x.cos^2x = sin2x - 1/2
<=> 1 - 1/2.(2sinx.cosx)^2 = sin2x - 1/2
<=> 2 - (2sinx.cosx)^2 = 2sin2x - 1
<=> 2 - sin^2(2x) = 2sin2x - 1
<=> sin^2(2x) + 2sin2x - 3 = 0
<=> (sin2x - 1)(sin2x + 3) = 0
<=> sin2x - 1 = 0 hoặc sin2x + 3 = 0, loại
<=> sin2x = 1
<=> 2x = π/2 + k2π, k nguyên
<=> x = π/4 + kπ, k nguyên