Share
Sin^4x + cos^4(x+π/4) =1/4
Question
Leave an answer
RuslanHeatt
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Answers ( )
Giải thích các bước giải:
[(1-cos2x)/2 ]^2 + [1+cos(2x+pi/2) /2 ]^2 =1/4
<=>(1-cos2x)^2 +[1-sin2x]^2=1
<=>1-2cos2x+cos^2 2x +1 -2sin2x+sin^2 2x=1
<=>1+sin^2 2x+cos^2 2x -2cos2x-2sin2x=0
<=> 2-2cos2x-2sin2x=0
<=>2cos2x+2sin2x=2
<=>2 sin(2x + pi/4)=2
<=> sin(2x+pi/4)=1
<=> 2x+pi/4= pi/2 + k2pi
<=>2x=pi/4+k2pi
<=>x=pi/8+kpi(kE Z)
Đáp án:
$x=\left({k\pi;\dfrac{\pi}4+k\pi}\right)$ $(k\in\mathbb Z)$
Lời giải:
$\sin^4x+\cos^4\left({x+\dfrac{\pi}4}\right)=\dfrac14$
$\Leftrightarrow\sin^4x+\left({\cos x.\dfrac1{\sqrt2}-\sin x.\dfrac1{\sqrt2}}\right)^4=\dfrac14$
$\Leftrightarrow\sin^4x+\dfrac14(\cos x-\sin x)^4=\dfrac14$
$\Leftrightarrow\sin^4x+\dfrac14[(\cos x-\sin x)^2]^2=\dfrac14$
$\Leftrightarrow\sin^4x+\dfrac14(\cos^2x-2\sin x\cos x+\sin^2x)^2=\dfrac14$
$\Leftrightarrow\sin^4x+\dfrac14(1-2\sin x\cos x)^2=\dfrac14$
$\Leftrightarrow\sin^4x+\sin^2x\cos^2x-\sin x\cos x=0$
$\left[\begin{array}{I}\sin x=0\text{ (1)}\\\sin^3x+\sin x\cos^2x-\cos x=0\text{ (2)}\end{array}\right.$
(1) $\Leftrightarrow x=k\pi$ $(k\in\mathbb Z)$
(2) $\Leftrightarrow \sin x(\sin^2x+\cos^2x)-\cos x=0$
$\Leftrightarrow\sin x-\cos x=0$
$\Leftrightarrow\sin\left({x-\dfrac{\pi}4}\right)=0$
$\Leftrightarrow x-\dfrac{\pi}4=k\pi$
$\Leftrightarrow x=\dfrac{\pi}4+k\pi$ $(k\in\mathbb Z)$
Vậy phương trình có nghiệm:
$x=\left({k\pi;\dfrac{\pi}4+k\pi}\right)$ $(k\in\mathbb Z)$.
Để tìm câu trả lời chính xác các em hãy tham khảo sin^4x các nguồn hoc24.vn, lazi.vn, hoidap247.com để thầy cô và các chuyên gia hỗ trợ các em nhé!