a) `cos(x+π)=1`
⇔ `x+π=k2π`
⇔ `x=-π+k2π \ (k∈\mathbb{Z})`
Vậy `x=-π+k2π \ (k∈\mathbb{Z})`
b) `sin(x+30^o)=sin(2x+55^o)`
⇔ $\left [\begin{array}{l} x+30^o=2x+55^o+k360^o \\ x+30^o=180^o-2x-55^o+k360^o \end{array} \right.$
⇔ $\left [\begin{array}{l} -x=25^o+k360^o \\ 3x=95^o+k360^o \end{array} \right.$
⇔ $\left [\begin{array}{l} x=-25^o-k360^o \\ x=\dfrac{95^o}{3}+k120^o \end{array} \right. \ (k∈\mathbb{Z})$
Vậy `x=-25^o-k360^o` hoặc `x=\frac{95^o}{3}+k120^o \ (k∈\mathbb{Z})`
