已知方程sinx 根号3cosx a=0

/>tanx=√3则x=kπ+π/3,k∈Z（1）k是偶数,则sinx=√3/2,cosx=1/2（2）k是奇数则sinx=√3/2,cosx=1/2(sinx+cosx)/(sinx-cosx)的值

用sinx2+cosx2=1再问：可以帮我把顺序写出吗。再问：谢谢。再答：就是3cos2x+cos2x=1再答：然后cosx=正负0.5就行了再问：就只有后面这两个步骤吧再问：还是前面那个也要再问：s

sinx+cosx=√2两边同时平方得sin²x+2sinxcosx+cos²x=21+2sinxcosx=22sinxcosx=1sinxcosx=1/2答案：1/2

F(x)=m.n=2根号3sinxcosx+cos^2x-sin^2x=根号3sin2x+cos2x=2sin(2x+π/6)

cosx-sinx=(3/5)*根号2,平方,1-2sinx*cosx=18/25可得2sinx*cosx=7/25,1+2sinx*cosx=32/25,即（sinx+cosx)^2=32/25因为

sinx+根号3cosx=22(sinx/2+根号3cosx/2)=2sinx/2+根号3cosx/2=1sinxcosπ/3+cosxsinπ/3=1sin(x+π/3)=1所以x+π/3=2kπ+

原式=（2sinxcosx-2sin^2x）/[（cosx-sinx）/cosx]=2sinxcosx（cosx-sinx）/（cosx-sinx）=2sinxcosx又（cosx-sinx）^2=c

解:由题意可得:(sinx+cosx)^2=1+sin2x=1-√3/2所以sin2x=-√3/2,所以2x=π+π/3或2x=2π-π/3因为0

f(x)=sinx+根号3cosx=2*sin(x+pi/3)1.T=2pi2.x用x-pi/3代替:y=sinx单调增区间：【0,pi/2】

=2*(根号3sinx/2+cosx/2)=2(sinx*cos30°+cosxsin30°）=2sin（x+30°)

我按根号下3/2,给你答案.两边平方得：(sinx)^2+2sinxcosx+(cosx)^2=7/4-√3sinxcosx=3/8-√3/2sinx和cosx看作是二元一次方程的两个根得方程X^2-

y=2Sin(x+π/3）对称轴是x=π/6+kπ,（k为整数）

1.f(x)=2(√3sinxcosx+(cosx)^2)+2m-1=√3sin2x+cos2x+2m=2sin(2x+pi/6)+2m最小正周期=pi2.x属于[0,pi/2]f(x)最小值=2si

(1)a*b=0sin2x-cos2x=0sqr(2)sin(2x-π/4)=0x=π/8+kπ/2,k∈Z(2)f(x)=sqr(2)sin(2x-π/4)x∈(3π/8+kπ,7π/8+kπ),k

f(x)=2(sinx*√3/2-cosx*1/2)=2(sinxcosπ/6-cosxsinπ/6)=2sin(x-π/6)、1/2表示2分之1=====sinx=4/5,题目应该有x∈〔-π/2,

可是我用sinxcosx=-根号3/4sinx=-根号3/4*cosx你确定移项过来不应该是除么.

f(x)=sinx+根号3cosx=2sin(x+π/3),即最小正周期为2π得到的g(x)=2sin(x+π/3-π/3)=2sinx,即在(O,π/2】上单调递增,在【π/2,π）上单调递减

.sinx+cosx=根号3/2,平方得:1+2sinxcosx=3/4则sin2x=3/4-1=-1/4

没错,f(x)=2sin(2x+π/6)周期T=2π/2=π因为-1≤sin(2x+π/6)≤1f(x)max=2f(x)min=-2

2√3(1-sin²x)=sinx2√3sin²x+sinx-2√3=0(2sinx-√3)(√3sinx+2)=02sinx-√3=0或√3sinx+2=0sinx=√3/2或s