1-sinx=(sinx+cosx)*cosx

Đáp án:

\(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+k2\pi\\x=\pi+k2\pi\end{array} \right.\) 

Giải thích các bước giải:

 $1-sinx=(sinx+cosx).cosx$

$1-sinx=sinx.cosx+cos^2x$

$sin^2x+cos^2x=sinx.cosx+sinx+cos^2x$

$sin^2x-sinx=sinx.cosx$

$sinx(sinx-1)=sinx.cosx$

$sinx-1=cosx$

$sinx-cosx=1$

ta có :

$a^2+b^2=2\geq 1$

Chia hai vế cho $\sqrt{a^2+b^2}=\sqrt{2}$

$\dfrac{1}{\sqrt{2}}sinx-\dfrac{1}{\sqrt{2}}cosx=\dfrac{1}{\sqrt{2}}$

$cos\dfrac{\pi}{4}sinx-sin\dfrac{\pi}{4}cosx=\dfrac{1}{\sqrt{2}}$
$sin(x-\dfrac{\pi}{4})=\dfrac{1}{\sqrt{2}}$

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

\(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+k2\pi\\x=\pi+k2\pi\end{array} \right.\)