Đáp án:

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

Lời giải:

Cách 1:

$\sin x = \cos x$

$\Leftrightarrow \sin x-\cos x=0$

$\Leftrightarrow \dfrac{1}{{\sqrt{2}}}\sin x -\dfrac{1}{{\sqrt{2}}}\cos x = 0$

$\Leftrightarrow \cos \dfrac{\pi}{4}.\sin x -\sin \dfrac{\pi}{4}.\cos x = 0$ 

$\Leftrightarrow\sin\left({x -\dfrac{\pi}{4}}\right) = 0$

$\Leftrightarrow x - \dfrac{\pi}{4}=k\pi$

$\Leftrightarrow x = \dfrac{\pi}{4}+k\pi\ (k\in\mathbb{Z})$

Vậy $x = \dfrac{\pi}{4}+k\pi\ (k\in\mathbb{Z})$

Cách 2:

$\sin x = \cos x$

$\Leftrightarrow \sqrt2\left(\sin x.\dfrac{\sqrt2}{2} - \cos x .\dfrac{\sqrt2}{2}\right)$

$\Leftrightarrow \sqrt2\left(\sin x.\cos \dfrac{\pi}{4} - \cos x. \sin \dfrac{\pi}{4} \right)=0$

$\Leftrightarrow \sqrt2\sin\left(x - \dfrac{\pi}{4}\right)=0$

$\Leftrightarrow \sin\left(x - \dfrac{\pi}{4}\right)=0$

$\Leftrightarrow x-\dfrac{\pi}{4}=k\pi\ (k\in\mathbb{Z})$

$\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\ (k\in\mathbb{Z})$

Đáp án:

Giải thích các bước giải:

 sinx=cosx⇔sinx=sin(pi/2-x)

⇔x=pi/2-x+k2pi hoặc x=pi-pi/2+x+k2pi⇒x=pi/4+kpi