三角函数方程问题

题目

三角函数方程问题
Solve the following equation for 0°≤x≤360°.(Give the answers correct to 1 decimal place if necessary.)
-2 tan ( 3x + 6° ) = 1

答案

-2tan(3x+6°)=1
tan(3x+6°)=-1/2
0°≤x≤360°
0°≤3x≤1080°
6°≤3x+6°≤1086°
The reference angle is 3x+6°=arctan(-1/2)≈-26.5651°
∵tan(3x+6°)<0
∴3x+6° lies in quadrant II or IV.
3x+6°=
180°-26.5651°=153.4°
360-26.5651°=333.4°
540-26.5651°=513.4°
720-26.5651°=693.4°
900-26.5651°=873.4°
1080-26.5651°=1053.4°
∴x=
49.1°
109.1°
169.1°
229.1°
289.1°
349.1°