最简三角方程题目√√√√√√√√√√√√(1)解方程:7-4sin^x-4√3cosx=0 (2)设方程x^+3√3x+4=0的两根为x1,x2,a=arctanx1,b=arctanx2,求a+b的值为什么这道题a+b的值域为(-pai,0)?注:以上^为平方,a,
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![最简三角方程题目√√√√√√√√√√√√(1)解方程:7-4sin^x-4√3cosx=0 (2)设方程x^+3√3x+4=0的两根为x1,x2,a=arctanx1,b=arctanx2,求a+b的值为什么这道题a+b的值域为(-pai,0)?注:以上^为平方,a,](/uploads/image/z/5245285-13-5.jpg?t=%E6%9C%80%E7%AE%80%E4%B8%89%E8%A7%92%E6%96%B9%E7%A8%8B%E9%A2%98%E7%9B%AE%E2%88%9A%E2%88%9A%E2%88%9A%E2%88%9A%E2%88%9A%E2%88%9A%E2%88%9A%E2%88%9A%E2%88%9A%E2%88%9A%E2%88%9A%E2%88%9A%EF%BC%881%EF%BC%89%E8%A7%A3%E6%96%B9%E7%A8%8B%EF%BC%9A7%EF%BC%8D4sin%5Ex-4%E2%88%9A3cosx%3D0+%EF%BC%882%EF%BC%89%E8%AE%BE%E6%96%B9%E7%A8%8Bx%5E%2B3%E2%88%9A3x%2B4%3D0%E7%9A%84%E4%B8%A4%E6%A0%B9%E4%B8%BAx1%2Cx2%2Ca%3Darctanx1%2Cb%3Darctanx2%2C%E6%B1%82a%2Bb%E7%9A%84%E5%80%BC%E4%B8%BA%E4%BB%80%E4%B9%88%E8%BF%99%E9%81%93%E9%A2%98a%2Bb%E7%9A%84%E5%80%BC%E5%9F%9F%E4%B8%BA%28-pai%2C0%29%3F%E6%B3%A8%EF%BC%9A%E4%BB%A5%E4%B8%8A%5E%E4%B8%BA%E5%B9%B3%E6%96%B9%2Ca%2C)
最简三角方程题目√√√√√√√√√√√√(1)解方程:7-4sin^x-4√3cosx=0 (2)设方程x^+3√3x+4=0的两根为x1,x2,a=arctanx1,b=arctanx2,求a+b的值为什么这道题a+b的值域为(-pai,0)?注:以上^为平方,a,
最简三角方程题目√√√√√√√√√√√√
(1)解方程:7-4sin^x-4√3cosx=0
(2)设方程x^+3√3x+4=0的两根为x1,x2,a=arctanx1,b=arctanx2,求a+b的值
为什么这道题a+b的值域为(-pai,0)?
注:以上^为平方,a,b为角阿尔法,和角贝塔.
最简三角方程题目√√√√√√√√√√√√(1)解方程:7-4sin^x-4√3cosx=0 (2)设方程x^+3√3x+4=0的两根为x1,x2,a=arctanx1,b=arctanx2,求a+b的值为什么这道题a+b的值域为(-pai,0)?注:以上^为平方,a,
1.x=2kπ+arcsin(7/8)-π/3
2.a+b=arctanx1+arctanx2
则tan(a+b)=tan(arctanx1+arctanx2)=x1+x2/(1-x1*x2)
x1+x2=-3√3,x1*x2=4
所以,tan(a+b)=√3,a+b=kπ+π/3
1.提出一个8来 :7-8*(1/2sin x -√3/2 cos x)=0
运用正弦和角公式:7-8*sin (x+π/3)=0
sin (x+π/3)=7/8
x=2kπ+arcsin(7/8)-π/3
2:a+b=arctanx1+arctanx2
则tan(a+b)=tan(arctanx1+arctanx2)=x1+x2/(1-x1*x2)
x1+x2=-3√3,x1*x2=4
补充:x1+x2<0,x1*x2>0 所以 -π