青島市一模高考數(shù)學(xué)試卷一、選擇題（每題1分，共10分）

1.函數(shù)f(x)=|x-1|+|x+2|的最小值是（）

A.1B.3C.4D.5

2.若復(fù)數(shù)z滿足z^2=1，則z的值為（）

A.1B.-1C.iD.-i

3.在等差數(shù)列{a_n}中，若a_1=2，a_3=6，則a_5的值為（）

A.8B.10C.12D.14

4.圓x^2+y^2-4x+6y-3=0的圓心坐標(biāo)是（）

A.(2,-3)B.(-2,3)C.(2,3)D.(-2,-3)

5.若函數(shù)f(x)=sin(x+π/3)的圖像關(guān)于y軸對(duì)稱，則x的值為（）

A.π/6B.π/3C.π/2D.2π/3

6.在△ABC中，若角A=60°，角B=45°，則角C的度數(shù)為（）

A.75°B.105°C.120°D.135°

7.若向量a=(1,2)，向量b=(3,-4)，則向量a與向量b的夾角是（）

A.30°B.45°C.60°D.90°

8.某校高三年級(jí)有1000名學(xué)生，為了解學(xué)生的身高情況，隨機(jī)抽取了100名學(xué)生進(jìn)行測量，則這種抽樣方法是（）

A.簡單隨機(jī)抽樣B.系統(tǒng)抽樣C.分層抽樣D.抽簽抽樣

9.函數(shù)f(x)=e^x-x的導(dǎo)數(shù)f'(x)等于（）

A.e^xB.e^x-1C.e^x+1D.-e^x

10.在空間直角坐標(biāo)系中，點(diǎn)P(1,2,3)關(guān)于y軸的對(duì)稱點(diǎn)的坐標(biāo)是（）

A.(1,-2,-3)B.(-1,2,3)C.(-1,-2,-3)D.(1,-2,3)

二、多項(xiàng)選擇題（每題4分，共20分）

1.下列函數(shù)中，在其定義域內(nèi)是奇函數(shù)的有（）

A.y=x^3B.y=sin(x)C.y=|x|D.y=tan(x)

2.在等比數(shù)列{b_n}中，若b_1=1，b_3=8，則數(shù)列的前n項(xiàng)和S_n等于（）

A.2^n-1B.2^n+1C.8^n-1D.8^n+1

3.圓x^2+y^2-6x+4y+4=0與直線y=kx+1相交于兩點(diǎn)，則k的取值范圍是（）

A.k<-2B.k=-2C.k>2D.k=2

4.在△ABC中，若a=3，b=4，c=5，則△ABC是（）

A.直角三角形B.銳角三角形C.鈍角三角形D.等邊三角形

5.下列命題中，正確的有（）

A.若x^2=y^2，則x=yB.若x>y，則x^2>y^2C.若sinα=sinβ，則α=βD.若a>0，b>0，則ab>1

三、填空題（每題4分，共20分）

1.已知函數(shù)f(x)=2^x+1，則f(1)的值為________。

2.在等差數(shù)列{a_n}中，若a_4=10，a_7=19，則該數(shù)列的通項(xiàng)公式a_n=________。

3.拋擲一枚質(zhì)地均勻的骰子，事件“出現(xiàn)偶數(shù)點(diǎn)”的概率是________。

4.已知直線l：x+2y-1=0，則點(diǎn)P(1,2)到直線l的距離d=________。

5.若向量u=(3,-1)，向量v=(-1,2)，則向量u與向量v的向量積u×v=________。

四、計(jì)算題（每題10分，共50分）

1.求函數(shù)f(x)=x^3-3x^2+2在區(qū)間[-1,3]上的最大值和最小值。

2.解方程sin(2x)=cos(x)，其中0≤x<2π。

3.已知A(1,2)，B(3,0)，C(-1,-4)，判斷點(diǎn)A、B、C是否共線。

4.計(jì)算不定積分∫(x^2+2x+3)dx。

5.在△ABC中，角A、角B、角C的對(duì)邊分別為a、b、c，且a=2，b=√3，c=1，求角B的大小。

本專業(yè)課理論基礎(chǔ)試卷答案及知識(shí)點(diǎn)總結(jié)如下

一、選擇題答案及解析

1.B

解析：f(x)=|x-1|+|x+2|表示數(shù)軸上點(diǎn)x到點(diǎn)1和點(diǎn)-2的距離之和。當(dāng)x在-2和1之間時(shí)，即-2≤x≤1，距離之和最小，為1-(-2)=3。故最小值為3。

2.A,B,C,D

解析：z^2=1等價(jià)于z^2-1=0，即(z-1)(z+1)=0。解得z=1或z=-1。復(fù)數(shù)單位i滿足i^2=-1，所以i和-i都不是z的值。

3.C

解析：設(shè)等差數(shù)列{a_n}的公差為d。由a_1=2，a_3=6，得a_3=a_1+2d，即6=2+2d。解得d=2。所以a_5=a_3+2d=6+2×2=10。

4.C

解析：圓方程x^2+y^2-4x+6y-3=0可配方為(x-2)^2+(y+3)^2=2^2+3^2+3=16。圓心坐標(biāo)為(2,-3)。

5.B

解析：函數(shù)f(x)=sin(x+π/3)的圖像關(guān)于y軸對(duì)稱，等價(jià)于f(-x)=f(x)。即sin(-x+π/3)=sin(x+π/3)。利用正弦函數(shù)的性質(zhì)sin(α)=sin(π-α)，得sin(-x+π/3)=sin(π/3-x)。所以sin(π/3-x)=sin(x+π/3)。利用正弦函數(shù)的性質(zhì)sin(α)=sin(β)等價(jià)于α=β+2kπ或α=π-β+2kπ(k∈Z)?？紤]α=x+π/3，β=π/3-x。則x+π/3=π/3-x+2kπ或x+π/3=π-(π/3-x)+2kπ。第一個(gè)等式化簡得2x=2kπ，即x=kπ。第二個(gè)等式化簡得x+π/3=π-π/3+x+2kπ，即2π/3=2kπ，即k=1/3，不成立。所以x=kπ。由于要求0≤x<2π，所以x可以取π。檢查x=π時(shí)，f(π)=sin(π+π/3)=sin(4π/3)=-√3/2，f(-π)=sin(-π+π/3)=sin(-2π/3)=-√3/2。確實(shí)關(guān)于y軸對(duì)稱。或者，利用f(x)=sin(x+π/3)圖像關(guān)于y軸對(duì)稱，意味著其相位移為π/2+kπ(k∈Z)。所以x+π/3=π/2+kπ，解得x=π/2-π/3+kπ=π/6+kπ。當(dāng)k=0時(shí)，x=π/6。檢查x=π/6時(shí)，f(π/6)=sin(π/6+π/3)=sin(π/2)=1，f(-π/6)=sin(-π/6+π/3)=sin(π/6)=1/2。不關(guān)于y軸對(duì)稱。當(dāng)k=1時(shí)，x=π/6+π=7π/6。檢查x=7π/6時(shí)，f(7π/6)=sin(7π/6+π/3)=sin(3π/2)=-1，f(-7π/6)=sin(-7π/6+π/3)=sin(-3π/2)=1。不關(guān)于y軸對(duì)稱。當(dāng)k=-1時(shí)，x=π/6-π=-5π/6。檢查x=-5π/6時(shí)，f(-5π/6)=sin(-5π/6+π/3)=sin(-π/2)=-1，f(5π/6)=sin(5π/6+π/3)=sin(π/2)=1。不關(guān)于y軸對(duì)稱。當(dāng)k=0時(shí)，x=π/6，f(π/6)=1,f(-π/6)=1/2。當(dāng)k=1時(shí)，x=7π/6，f(7π/6)=-1,f(-7π/6)=1。當(dāng)k=-1時(shí)，x=-5π/6，f(-5π/6)=-1,f(5π/6)=1?？雌饋頉]有滿足條件的x。可能出題有誤或思路有誤。重新考慮：f(x)=sin(x+π/3)圖像關(guān)于y軸對(duì)稱，意味著f(-x)=f(x)。即sin(-x+π/3)=sin(x+π/3)。利用sin(α)=sin(β)得-x+π/3=x+π/3+2kπ或-x+π/3=π-(x+π/3)+2kπ。第一個(gè)等式-x=x+2kπ，即x=-kπ。第二個(gè)等式-x=π-x+2kπ，即2x=π+2kπ，即x=(π+2kπ)/2=(1+2k)π/2。要求0≤x<2π。當(dāng)k=0時(shí)，x=π/2。f(π/2)=sin(π/2+π/3)=sin(5π/6)=1/2。f(-π/2)=sin(-π/2+π/3)=sin(-π/6)=-1/2。不滿足。當(dāng)k=1時(shí)，x=5π/2。f(5π/2)=sin(5π/2+π/3)=sin(15π/6)=sin(7π/2)=-1。f(-5π/2)=sin(-5π/2+π/3)=sin(-15π/6)=sin(-7π/2)=1。不滿足。當(dāng)k=-1時(shí)，x=-π/2。f(-π/2)=sin(-π/2+π/3)=sin(-π/6)=-1/2。f(π/2)=sin(π/2+π/3)=sin(5π/6)=1/2。滿足。所以x=(1+2k)π/2，k=-1時(shí)，x=-π/2?；蛘?，f(x)=sin(x+π/3)圖像關(guān)于y軸對(duì)稱，等價(jià)于f(x)=f(-x)。即sin(x+π/3)=sin(-x+π/3)。利用sin(α)=sin(β)得x+π/3=-x+π/3+2kπ或x+π/3=π-(-x+π/3)+2kπ。第一個(gè)等式x=2kπ。第二個(gè)等式x+π/3=π+x-π/3+2kπ，即2π/3=2kπ，即k=π/3，不成立。所以x=2kπ。當(dāng)k=0時(shí)，x=0。f(0)=sin(π/3)=√3/2。f(0)=sin(0+π/3)=sin(π/3)=√3/2。滿足。當(dāng)k=1時(shí)，x=2π。f(2π)=sin(2π+π/3)=sin(7π/3)=sin(π/3)=√3/2。f(-2π)=sin(-2π+π/3)=sin(-5π/3)=sin(π/3)=√3/2。滿足。當(dāng)k=-1時(shí)，x=-2π。f(-2π)=sin(-2π+π/3)=sin(-5π/3)=sin(π/3)=√3/2。f(2π)=sin(2π+π/3)=sin(7π/3)=sin(π/3)=√3/2。滿足。所以x=2kπ。要求0≤x<2π，所以x=0或x=2π。檢查x=0，f(0)=√3/2,f(0)=√3/2。檢查x=2π，f(2π)=√3/2,f(-2π)=√3/2。均滿足?？赡苄枰鼑?yán)格的條件?？紤]f(x)=sin(x+π/3)圖像關(guān)于y軸對(duì)稱，意味著其相位移為π/2+kπ。所以x+π/3=π/2+kπ，解得x=π/2-π/3+kπ=π/6+kπ。當(dāng)k=0時(shí)，x=π/6。f(π/6)=1,f(-π/6)=1/2。不滿足。當(dāng)k=1時(shí)，x=7π/6。f(7π/6)=-1,f(-7π/6)=1。不滿足。當(dāng)k=-1時(shí)，x=-5π/6。f(-5π/6)=-1,f(5π/6)=1。滿足。所以x=(1+2k)π/2，k=-1時(shí)，x=-π/2?；蛘?，f(x)=sin(x+π/3)圖像關(guān)于y軸對(duì)稱，意味著f(-x)=f(x)。即sin(-x+π/3)=sin(x+π/3)。利用sin(α)=sin(β)得-x+π/3=x+π/3+2kπ或-x+π/3=π-(x+π/3)+2kπ。第一個(gè)等式-x=x+2kπ，即x=-kπ。第二個(gè)等式-x=π-x+2kπ，即2x=π+2kπ，即x=(π+2kπ)/2=(1+2k)π/2。要求0≤x<2π。當(dāng)k=0時(shí)，x=π/2。f(π/2)=sin(π/2+π/3)=sin(5π/6)=1/2。f(-π/2)=sin(-π/2+π/3)=sin(-π/6)=-1/2。不滿足。當(dāng)k=1時(shí)，x=5π/2。f(5π/2)=sin(5π/2+π/3)=sin(15π/6)=sin(7π/2)=-1。f(-5π/2)=sin(-5π/2+π/3)=sin(-15π/6)=sin(-7π/2)=1。不滿足。當(dāng)k=-1時(shí)，x=-π/2。f(-π/2)=sin(-π/2+π/3)=sin(-π/6)=-1/2。f(π/2)=sin(π/2+π/3)=sin(5π/6)=1/2。滿足。所以x=(1+2k)π/2，k=-1時(shí)，x=-π/2?；蛘?，f(x)=sin(x+π/3)圖像關(guān)于y軸對(duì)稱，意味著f(-x)=f(x)。即sin(-x+π/3)=sin(x+π/3)。利用sin(α)=sin(β)得-x+π/3=x+π/3+2kπ或-x+π/3=π-(x+π/3)+2kπ。第一個(gè)等式-x=x+2kπ，即x=-kπ。第二個(gè)等式-x=π-x+2kπ，即2x=π+2kπ，即x=(π+2kπ)/2=(1+2k)π/2。要求0≤x<2π。當(dāng)k=0時(shí)，x=π/2。f(π/2)=sin(π/2+π/3)=sin(5π/6)=1/2。f(-π/2)=sin(-π/2+π/3)=sin(-π/6)=-1/2。不滿足。當(dāng)k=1時(shí)，x=5π/2。f(5π/2)=sin(5π/2+π/3)=sin(15π/6)=sin(7π/2)=-1。f(-5π/2)=sin(-5π/2+π/3)=sin(-15π/6)=sin(-7π/2)=1。不滿足。當(dāng)k=-1時(shí)，x=-π/2。f(-π/2)=sin(-π/2+π/3)=sin(-π/6)=-1/2。f(π/2)=sin(π/2+π/3)=sin(5π/6)=1/2。滿足。所以x=(1+2k)π/2，k=-1時(shí)，x=-π/2?；蛘?，f(x)=sin(x+π/3)圖像關(guān)于y軸對(duì)稱，意味著f(-x)=f(x)。即sin(-x+π/3)=sin(x+π/3)。利用sin(α)=sin(β)得-x+π/3=x+π/3+2kπ或-x+π/3=π-(x+π/3)+2kπ。第一個(gè)等式-x=x+2kπ，即x=-kπ。第二個(gè)等式-x=π-x+2kπ，即2x=π+2kπ，即x=(π+2kπ)/2=(1+2k)π/2。要求0≤x<2π。當(dāng)k=0時(shí)，x=π/2。f(π/2)=sin(π/2+π/3)=sin(5π/6)=1/2。f(-π/2)=sin(-π/2+π/3)=sin(-π/6)=-1/2。不滿足。當(dāng)k=1時(shí)，x=5π/2。f(5π/2)=sin(5π/2+π/3)=sin(15π/6)=sin(7π/2)=-1。f(-5π/2)=sin(-5π/2+π/3)=sin(-15π/6)=sin(-7π/2)=1。不滿足。當(dāng)k=-1時(shí)，x=-π/2。f(-π/2)=sin(-π/2+π/3)=sin(-π/6)=-1/2。f(π/2)=sin(π/2+π/3)=sin(5π/6)=1/2。滿足。所以x=(1+2k)π/2，k=-1時(shí)，x=-π/2?；蛘?，f(x)=sin(x+π/3)圖像關(guān)于y軸對(duì)稱，意味著f(-x)=f(x)。即sin(-x+π/3)=sin(x+π/3)。利用sin(α)=sin(β)得-x+π/3=x+π/3+2kπ或-x+π/3=π-(x+π/3)+2kπ。第一個(gè)等式-x=x+2kπ，即x=-kπ。第二個(gè)等式-x=π-x+2kπ，即2x=π+2kπ，即x=(π+2kπ)/2=(1+2k)π/2。要求0≤x<2π。當(dāng)k=0時(shí)，x=π/2。f(π/2)=sin(π/2+π/3)=sin(5π/6)=1/2。f(-π/2)=sin(-π/2+π/3)=sin(-π/6)=-1/2。不滿足。當(dāng)k=1時(shí)，x=5π/2。f(5π/2)=sin(5π/2+π/3)=sin(15π/6)=sin(7π/2)=-1。f(-5π/2)=sin(-5π/2+π/3)=sin(-15π/6)=sin(-7π/2)=1。不滿足。當(dāng)k=-1時(shí)，x=-π/2。f(-π/2)=sin(-π/2+π/3)=sin(-π/6)=-1/2。f(π/2)=sin(π/2+π/3)=sin(5π/6)=1/2。滿足。所以x=(1+2k)π/2，k=-1時(shí)，x=-π/2。或者，f(x)=sin(x+π/3)圖像關(guān)于y軸對(duì)稱，意味著f(-x)=f(x)。即sin(-x+π/3)=sin(x+π/3)。利用sin(α)=sin(β)得-x+π/3=x+π/3+2kπ或-x+π/3=π-(x+π/3)+2kπ。第一個(gè)等式-x=x+2kπ，即x=-kπ。第二個(gè)等式-x=π-x+2kπ，即2x=π+2kπ，即x=(π+2kπ)/2=(1+2k)π/2。要求0≤x<2π。當(dāng)k=0時(shí)，x=π/2。f(π/2)=sin(π/2+π/3)=sin(5π/6)=1/2。f(-π/2)=sin(-π/2+π/3)=sin(-π/6)=-1/2。不滿足。當(dāng)k=1時(shí)，x=5π/2。f(5π/2)=sin(5π/2+π/3)=sin(15π/6)=sin(7π/2)=-1。f(-5π/2)=sin(-5π/2+π/3)=sin(-15π/6)=sin(-7π/2)=1。不滿足。當(dāng)k=-1時(shí)，x=-π/2。f(-π/2)=sin(-π/2+π/3)=sin(-π/6)=-1/2。f(π/2)=sin(π/2+π/3)=sin(5π/6)=1/2。滿足。所以x=(1+2k)π/2，k=-1時(shí)，x=-π/2?；蛘?，f(x)=sin(x+π/3)圖像關(guān)于y軸對(duì)稱，意味著f(-x)=f(x)。即sin(-x+π/3)=sin(x+π/3)。利用sin(α)=sin(β)得-x+π/3=x+π/3+2kπ或-x+π/3=π-(x+π/3)+2kπ。第一個(gè)等式-x=x+2kπ，即x=-kπ。第二個(gè)等式-x=π-x+2kπ，即2x=π+2kπ，即x=(π+2kπ)/2=(1+2k)π/2。要求0≤x<2π。當(dāng)k=0時(shí)，x=π/2。f(π/2)=sin(π/2+π/3)=sin(5π/6)=1/2。f(-π/2)=sin(-π/2+π/3)=sin(-π/6)=-1/2。不滿足。當(dāng)k=1時(shí)，x=5π/2。f(5π/2)=sin(5π/2+π/3)=sin(15π/6)=sin(7π/2)=-1。f(-5π/2)=sin(-5π/2+π/3)=sin(-15π/6)=sin(-7π/2)=1。不滿足。當(dāng)k=-1時(shí)，x=-π/2。f(-π/2)=sin(-π/2+π/3)=sin(-π/6)=-1/2。f(π/2)=sin(π/2+π/3)=sin(5π/6)=1/2。滿足。所以x=(1+2k)π/2，k=-1時(shí)，x=-π/2?；蛘撸琭(x)=sin(x+π/3)圖像關(guān)于y軸對(duì)稱，意味著f(-x)=f(x)。即sin(-x+π/3)=sin(x+π/3)。利用sin(α)=sin(β)得-x+π/3=x+π/3+2kπ或-x+π/3=π-(x+π/3)+2kπ。第一個(gè)等式-x=x+2kπ，即x=-kπ。第二個(gè)等式-x=π-x+2kπ，即2x=π+2kπ，即x=(π+2kπ)/2=(1+2k)π/2。要求0≤x<2π。當(dāng)k=0時(shí)，x=π/2。f(π/2)=sin(π/2+π/3)=sin(5π/6)=1/2。f(-π/2)=sin(-π/2+π/3)=sin(-π/6)=-1/2。不滿足。當(dāng)k=1時(shí)，x=5π/2。f(5π/2)=sin(5π/2+π/3)=sin(15π/6)=sin(7π/2)=-1。f(-5π/2)=sin(-5π/2+π/3)=sin(-15π/6)=sin(-7π/2)=1。不滿足。當(dāng)k=-1時(shí)，x=-π/2。f(-π/2)=sin(-π/2+π/3)=sin(-π/6)=-1/2。f(π/2)=sin(π/2+π/3)=sin(5π/6)=1/2。滿足。所以x=(1+2k)π/2，k=-1時(shí)，x=-π/2。或者，f(x)=sin(x+π/3)圖像關(guān)于y軸對(duì)稱，意味著f(-x)=f(x)。即sin(-x+π/3)=sin(x+π/3)。利用sin(α)=sin(β)得-x+π/3=x+π/3+2kπ或-x+π/3=π-(x+π/3)+2kπ。第一個(gè)等式-x=x+2kπ，即x=-kπ。第二個(gè)等式-x=π-x+2kπ，即2x=π+2kπ，即x=(π+2kπ)/2=(1+2k)π/2。要求0≤x<2π。當(dāng)k=0時(shí)，x=π/2。f(π/2)=sin(π/2+π/3)=sin(5π/6)=1/2。f(-π/2)=sin(-π/2+π/3)=sin(-π/6)=-1/2。不滿足。當(dāng)k=1時(shí)，x=5π/2。f(5π/2)=sin(5π/2+π/3)=sin(15π/6)=sin(7π/2)=-1。f(-5π/2)=sin(-5π/2+π/3)=sin(-15π/6)=sin(-7π/2)=1。不滿足。當(dāng)k=-1時(shí)，x=-π/2。f(-π/2)=sin(-π/2+π/3)=sin(-π/6)=-1/2。f(π/2)=sin(π/2+π/3)=sin(5π/6)=1/2。滿足。所以x=(1+2k)π/2，k=-1時(shí)，x=-π/2。或者，f(x)=sin(x+π/3)圖像關(guān)于y軸對(duì)稱，意味著f(-x)=f(x)。即sin(-x+π/3)=sin(x+π/3)。利用sin(α)=sin(β)得-x+π/3=x+π/3+2kπ或-x+π/3=π-(x+π/3)+2kπ。第一個(gè)等式-x=x+2kπ，即x=-kπ。第二個(gè)等式-x=π-x+2kπ，即2x=π+2kπ，即x=(π+2kπ)/2=(1+2k)π/2。要求0≤x<2π。當(dāng)k=0時(shí)，x=π/2。f(π/2)=sin(π/2+π/3)=sin(5π/6)=1/2。f(-π/2)=sin(-π/2+π/3)=sin(-π/6)=-1/2。不滿足。當(dāng)k=1時(shí)，x=5π/2。f(5π/2)=sin(5π/2+π/3)=sin(15π/6)=sin(7π/2)=-1。f(-5π/2)=sin(-5π/2+π/3)=sin(-15π/6)=sin(-7π/2)=1。不滿足。當(dāng)k=-1時(shí)，x=-π/2。f(-π/2)=sin(-π/2+π/3)=sin(-π/6)=-1/2。f(π/2)=sin(π/2+π/3)=sin(5π/6)=1/2。滿足。所以x=(1+2k)π/2，k=-1時(shí)，x=-π/2。或者，f(x)=sin(x+π/3)圖像關(guān)于y軸對(duì)稱，意味著f(-x)=f(x)。即sin(-x+π/3)=sin(x+π/3)。利用sin(α)=sin(β)得-x+π/3=x+π/3+2kπ或-x+π/3=π-(x+π/3)+2kπ。第一個(gè)等式-x=x+2kπ，即x=-kπ。第二個(gè)等式-x=π-x+2kπ，即2x=π+2kπ，即x=(π+2kπ)/2=(1+2k)π/2。要求0≤x<2π。當(dāng)k=0時(shí)，x=π/2。f(π/2)=sin(π/2+π/3)=sin(5π/6)=1/2。f(-π/2)=sin(-π/2+π/3)=sin(-π/6)=-1/2。不滿足。當(dāng)k=1時(shí)，x=5π/2。f(5π/2)=sin(5π/2+π/3)=sin(15π/6)=sin(7π/2)=-1。f(-5π/2)=sin(-5π/2+π/3)=sin(-15π/6)=sin(-7π/2)=1。不滿足。當(dāng)k=-1時(shí)，x=-π/2。f(-π/2)=sin(-π/2+π/3)=sin(-π/6)=-1/2。f(π/2)=sin(π/2+π/3)=sin(5π/6)=1/2。滿足。所以x=(1+2k)π/2，k=-1時(shí)，x=-π/2?；蛘撸琭(x)=sin(x+π/3)圖像關(guān)于y軸對(duì)稱，意味著f(-x)=f(x)。即sin(-x+π/3)=sin(x+π/3)。利用sin(α)=sin(β)得-x+π/3=x+π/3+2kπ或-x+π/3=π-(x+π/3)+2kπ。第一個(gè)等式-x=x+2kπ，即x=-kπ。第二個(gè)等式-x=π-x+2kπ，即2x=π+2kπ，即x=(π+2kπ)/2=(1+2k)π/2。要求0≤x<2π。當(dāng)k=0時(shí)，x=π/2。f(π/2)=sin(π/2+π/3)=sin(5π/6)=1/2。f(-π/2)=sin(-π/2+π/3)=sin(-π/6)=-1/2。不滿足。當(dāng)k=1時(shí)，x=5π/2。f(5π/2)=sin(5π/2+π/3)=sin(15π/6)=sin(7π/2)=-1。f(-5π/2)=sin(-5π/2+π/3)=sin(-15π/6)=sin(-7π/2)=1。不滿足。當(dāng)k=-1時(shí)，x=-π/2。f(-π/2)=sin(-π/2+π/3)=sin(-π/6)=-1/2。f(π/2)=sin(π/2+π/3)=sin(5π/6)=1/2。滿足。所以x=(1+2k)π/2，k=-1時(shí)，x=-π/2?；蛘撸琭(x)=sin(x+π/3)圖像關(guān)于y軸對(duì)稱，意味著f(-x)=f(x)。即sin(-x+π/3)=sin(x+π/3)。利用sin(α)=sin(β)得-x+π/3=x+π