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Mk cần lời giải chi tiết

Tham khảo :

$3cos(x)+4sin(x)=6-\dfrac{6}{3cos(x)+4sin(x)+1}$$(1)$

`-` Đặt $t=3cos(x)+4sin(x)$$(-5 \leq t \leq 5;t \ne -1)$

$(1)$ trở thành: $t=6-\dfrac{6}{t+1}$

`<=>`$t(t+1)=6(t+1)-6$

`<=>`$t^2+t=6t+6-6$

`<=>`$t^2-5t=0$

`<=>`$t(t-5)=0$

`<=>`$\left[\begin{matrix} t=0(thỏa)\\ t=5(thỏa)\end{matrix}\right.$

`@` Khi $t=0,(x\ne \dfrac{\pi }{2}+2\pi k,\:x \ne \dfrac{3\pi }{2}+2\pi k)$

$3cos(x)+4sin(x)=0$

`<=>`$3+4\dfrac{sin(x)}{cos(x)}=0$

`<=>`$3+4tan(x)=0$

`<=>`$tan(x)=\dfrac{-3}{4}$

`<=>`$x=arctan(\dfrac{-3}{4})+k \pi(thỏa)$

`@` Khi $t=5$

$3cos(x)+4sin(x)=5$ $(a=3;b=4$`->`$\sqrt{3^2+4^2}=5)$

`<=>`$\dfrac{3}{5}cos(x)+\dfrac{4}{5}sin(x)=1$

`<=>`$sin[arcsin(\dfrac{3}{5})]cos(x)+sin(x)cos[arcsin(\dfrac{4}{5})]=sin(\dfrac{\pi }{2})$

`<=>`$sin[x+arcsin(\dfrac{3}{5})]=sin(\dfrac{\pi }{2})$

`<=>`$\left[\begin{matrix} x+arcsin(\dfrac{3}{5})=\dfrac{\pi}{2}+k2 \pi\\ x+arcsin(\dfrac{3}{5})=\pi -\dfrac{\pi}{2}+k2 \pi\end{matrix}\right.$

`<=>`$\left[\begin{matrix} x=\dfrac{\pi}{2}-arcsin(\dfrac{3}{5})+k2 \pi\\ x=\dfrac{\pi}{2}-arcsin(\dfrac{3}{5})+k2 \pi\end{matrix}\right.$

`<=>`$x=\dfrac{\pi}{2}-arcsin(\dfrac{3}{5})+k2 \pi$

Vậy phương trình có $2$ nghiệm là : 

$x=arctan(\dfrac{-3}{4})+k \pi$ hoặc $x=\dfrac{\pi}{2}-arcsin(\dfrac{3}{5})+k2 \pi$