三角方程2cos(x-π/4)=1在[0,4π]內的所有解的和是____( π)似乎很簡單哦可是我就是不會算不清楚了...幫幫忙...要過程..謝謝

熱心網友

2cos(x-π/4)=1得cos(x-π/4)=1/2所以x-π/4=2kπ+π/3 或2kπ-π/3 (k為整數)所以x=2kπ+7π/12 或2kπ-π/12 (k為整數)因為x在[0,4π],所以x=7π/12 或 23π/12 或 31π/12 或 47π/12所以7π/12 + 23π/12 + 31π/12 + 47π/12 =9π

熱心網友

化為cos(x-π/4)=1/2令t=x-π/4然后畫出在[0,4π]，cost的函數圖像可以知道 x-π/4可以等于π/3,5π/3,7π/3,11π/3解除四個x相加得9π

熱心網友

2cos(x-π/4)=1在 [0,4π]