Question

tan (2x+\(\dfrac{\Pi}{4}\)).tan(\(\Pi\)-\(\dfrac{x}{2}\))=1

Answer 1

Bài tập này áp dụng công thức phụ - chéo:

cot(a)=tan(\(\dfrac{\Pi}{2}\)-a) (cái này chắc bạn không quên chứ hihi)

Điều kiện: cos(2x+\(\dfrac{\Pi}{4}\))\(\ne\)0<=>x\(\ne\)\(\dfrac{\Pi}{8}\)+\(\dfrac{k\Pi}{2}\)

cos(\(\Pi\)-\(\dfrac{x}{2}\))\(\ne\)0<=>x\(\ne\)\(\Pi\)-2\(\Pi\)

PT<=>tan(2x+\(\dfrac{\Pi}{4}\))=\(\dfrac{1}{tan\left(\Pi-\dfrac{x}{2}\right)}\)

<=>tan(2x+\(\dfrac{\Pi}{4}\))=cot(\(\Pi\)-\(\dfrac{x}{2}\))

<=>tan(2x+\(\dfrac{\Pi}{4}\))=tan(\(\dfrac{\Pi}{2}\)-\(\Pi\)+\(\dfrac{x}{2}\))

<=>2x+\(\dfrac{\Pi}{4}\)=\(\dfrac{\Pi}{2}\)-\(\Pi\)+\(\dfrac{x}{2}\)

<=>x=-\(\dfrac{\Pi}{2}\)+k\(\dfrac{2\Pi}{3}\)(k\(\in\)Z)

Chúc bạn học tốt. Thân!

\(\dfrac{\Pi}{4}\)\(\Pi\)\(\Pi\)