sin⁴(x+π/4)=1/4+cos²x-cos⁴x

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1. Đáp án:

    $x = k\dfrac{\pi}{2}\quad (k\in \Bbb Z)$

   Giải thích các bước giải:

   $\sin^4\left(x + \dfrac{\pi}{4}\right) = \dfrac{1}{4} + \cos^2x – \cos^4x$

   $\Leftrightarrow \dfrac{1}{4}.\left[\sqrt2\sin\left(x + \dfrac{\pi}{4}\right)\right]^4= \dfrac{1}{4} + \cos^2x(1 – \cos^2x)$

   $\Leftrightarrow \dfrac{1}{4}(\sin x + \cos x)^4 = \dfrac{1}{4} + \cos^2x\sin^2x$

   Đặt $t = \sin x + \cos x \qquad (|t|\leq \sqrt2)$

   $\Rightarrow t^2 = 1 + 2\sin x\cos x$

   $\Rightarrow \dfrac{t^2 -1}{2} = \sin x\cos x$

   $\Rightarrow \left(\dfrac{t^2 -1}{2}\right)^2 = \sin^2x\cos^2x$

   Phương trình trở thành:

   $\dfrac{1}{4}t^4 = \dfrac{1}{4} + \left(\dfrac{t^2 -1}{2}\right)^2$

   $\Leftrightarrow t^4 = 1 + (t^2 – 1)^2$

   $\Leftrightarrow t^2 = 1$

   $\Leftrightarrow t = \pm 1$

   $+)$ Với $t = 1$ ta được:

   $\sin x + \cos x = 1$

   $\Leftrightarrow \sqrt2\left(x + \dfrac{\pi}{4}\right) = 1$

   $\Leftrightarrow \left(x + \dfrac{\pi}{4}\right) = \dfrac{\sqrt2}{2}$

   $\Leftrightarrow \left[\begin{array}{l}x + \dfrac{\pi}{4} = \dfrac{\pi}{4} + k2\pi\\x + \dfrac{\pi}{4} = \dfrac{3\pi}{4} + k2\pi\end{array}\right.$

   $\Leftrightarrow \left[\begin{array}{l}x = k2\pi\\x = \dfrac{\pi}{2} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$

   $+)$ Với $t = -1$ ta được:

   $\sin x + \cos x =- 1$

   $\Leftrightarrow \sqrt2\left(x + \dfrac{\pi}{4}\right) = -1$

   $\Leftrightarrow \left(x + \dfrac{\pi}{4}\right) = -\dfrac{\sqrt2}{2}$

   $\Leftrightarrow \left[\begin{array}{l}x + \dfrac{\pi}{4} = -\dfrac{\pi}{4} + k2\pi\\x + \dfrac{\pi}{4} = \dfrac{5\pi}{4} + k2\pi\end{array}\right.$

   $\Leftrightarrow \left[\begin{array}{l}x = -\dfrac{\pi}{2} + k2\pi\\x = \pi+ k2\pi\end{array}\right.\quad (k\in\Bbb Z)$

   Vậy phương trình đã cho có họ nghiệm:

   $x = k\dfrac{\pi}{2}\quad (k\in \Bbb Z)$

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