Sin^4x + cos^4(x+π/4) =1/4

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:

$x=\left({k\pi;\dfrac{\pi}4+k\pi}\right)$ $(k\in\mathbb Z)$.