解答
−11cos(x)22−11sin(x)8=−250
解答
x=0.10677…+2πn,x=0.59076…+2πn
+1
度数
x=6.11761…∘+360∘n,x=33.84860…∘+360∘n求解步骤
−11cos(x)⋅22−11sin(x)⋅8=−250
两边加上 11sin(x)8−242cos(x)=−250+88sin(x)
两边进行平方(−242cos(x))2=(−250+88sin(x))2
两边减去 (−250+88sin(x))258564cos2(x)−62500+44000sin(x)−7744sin2(x)=0
使用三角恒等式改写
−62500+44000sin(x)+58564cos2(x)−7744sin2(x)
使用毕达哥拉斯恒等式: cos2(x)+sin2(x)=1cos2(x)=1−sin2(x)=−62500+44000sin(x)+58564(1−sin2(x))−7744sin2(x)
化简 −62500+44000sin(x)+58564(1−sin2(x))−7744sin2(x):44000sin(x)−66308sin2(x)−3936
−62500+44000sin(x)+58564(1−sin2(x))−7744sin2(x)
乘开 58564(1−sin2(x)):58564−58564sin2(x)
58564(1−sin2(x))
使用分配律: a(b−c)=ab−aca=58564,b=1,c=sin2(x)=58564⋅1−58564sin2(x)
数字相乘:58564⋅1=58564=58564−58564sin2(x)
=−62500+44000sin(x)+58564−58564sin2(x)−7744sin2(x)
化简 −62500+44000sin(x)+58564−58564sin2(x)−7744sin2(x):44000sin(x)−66308sin2(x)−3936
−62500+44000sin(x)+58564−58564sin2(x)−7744sin2(x)
同类项相加:−58564sin2(x)−7744sin2(x)=−66308sin2(x)=−62500+44000sin(x)+58564−66308sin2(x)
对同类项分组=44000sin(x)−66308sin2(x)−62500+58564
数字相加/相减:−62500+58564=−3936=44000sin(x)−66308sin2(x)−3936
=44000sin(x)−66308sin2(x)−3936
=44000sin(x)−66308sin2(x)−3936
−3936+44000sin(x)−66308sin2(x)=0
用替代法求解
−3936+44000sin(x)−66308sin2(x)=0
令:sin(x)=u−3936+44000u−66308u2=0
−3936+44000u−66308u2=0:u=132616−440002−1043953152+44000,u=132616440002−1043953152+44000
−3936+44000u−66308u2=0
改写成标准形式 ax2+bx+c=0−66308u2+44000u−3936=0
使用求根公式求解
−66308u2+44000u−3936=0
二次方程求根公式:
若 a=−66308,b=44000,c=−3936u1,2=2(−66308)−44000±440002−4(−66308)(−3936)
u1,2=2(−66308)−44000±440002−4(−66308)(−3936)
440002−4(−66308)(−3936)=440002−1043953152
440002−4(−66308)(−3936)
使用法则 −(−a)=a=440002−4⋅66308⋅3936
数字相乘:4⋅66308⋅3936=1043953152=440002−1043953152
u1,2=2(−66308)−44000±440002−1043953152
将解分隔开u1=2(−66308)−44000+440002−1043953152,u2=2(−66308)−44000−440002−1043953152
u=2(−66308)−44000+440002−1043953152:132616−440002−1043953152+44000
2(−66308)−44000+440002−1043953152
去除括号: (−a)=−a=−2⋅66308−44000+440002−1043953152
数字相乘:2⋅66308=132616=−132616−44000+440002−1043953152
使用分式法则: −b−a=ba−44000+440002−1043953152=−(−440002−1043953152+44000)=132616−440002−1043953152+44000
u=2(−66308)−44000−440002−1043953152:132616440002−1043953152+44000
2(−66308)−44000−440002−1043953152
去除括号: (−a)=−a=−2⋅66308−44000−440002−1043953152
数字相乘:2⋅66308=132616=−132616−44000−440002−1043953152
使用分式法则: −b−a=ba−44000−440002−1043953152=−(440002−1043953152+44000)=132616440002−1043953152+44000
二次方程组的解是:u=132616−440002−1043953152+44000,u=132616440002−1043953152+44000
u=sin(x)代回sin(x)=132616−440002−1043953152+44000,sin(x)=132616440002−1043953152+44000
sin(x)=132616−440002−1043953152+44000,sin(x)=132616440002−1043953152+44000
sin(x)=132616−440002−1043953152+44000:x=arcsin(132616−440002−1043953152+44000)+2πn,x=π−arcsin(132616−440002−1043953152+44000)+2πn
sin(x)=132616−440002−1043953152+44000
使用反三角函数性质
sin(x)=132616−440002−1043953152+44000
sin(x)=132616−440002−1043953152+44000的通解sin(x)=a⇒x=arcsin(a)+2πn,x=π−arcsin(a)+2πnx=arcsin(132616−440002−1043953152+44000)+2πn,x=π−arcsin(132616−440002−1043953152+44000)+2πn
x=arcsin(132616−440002−1043953152+44000)+2πn,x=π−arcsin(132616−440002−1043953152+44000)+2πn
sin(x)=132616440002−1043953152+44000:x=arcsin(132616440002−1043953152+44000)+2πn,x=π−arcsin(132616440002−1043953152+44000)+2πn
sin(x)=132616440002−1043953152+44000
使用反三角函数性质
sin(x)=132616440002−1043953152+44000
sin(x)=132616440002−1043953152+44000的通解sin(x)=a⇒x=arcsin(a)+2πn,x=π−arcsin(a)+2πnx=arcsin(132616440002−1043953152+44000)+2πn,x=π−arcsin(132616440002−1043953152+44000)+2πn
x=arcsin(132616440002−1043953152+44000)+2πn,x=π−arcsin(132616440002−1043953152+44000)+2πn
合并所有解x=arcsin(132616−440002−1043953152+44000)+2πn,x=π−arcsin(132616−440002−1043953152+44000)+2πn,x=arcsin(132616440002−1043953152+44000)+2πn,x=π−arcsin(132616440002−1043953152+44000)+2πn
将解代入原方程进行验证
将它们代入 −11cos(x)22−11sin(x)8=−250检验解是否符合
去除与方程不符的解。
检验 arcsin(132616−440002−1043953152+44000)+2πn的解:真
arcsin(132616−440002−1043953152+44000)+2πn
代入 n=1arcsin(132616−440002−1043953152+44000)+2π1
对于 −11cos(x)22−11sin(x)8=−250代入x=arcsin(132616−440002−1043953152+44000)+2π1−11cos(arcsin(132616−440002−1043953152+44000)+2π1)⋅22−11sin(arcsin(132616−440002−1043953152+44000)+2π1)⋅8=−250
整理后得−250=−250
⇒真
检验 π−arcsin(132616−440002−1043953152+44000)+2πn的解:假
π−arcsin(132616−440002−1043953152+44000)+2πn
代入 n=1π−arcsin(132616−440002−1043953152+44000)+2π1
对于 −11cos(x)22−11sin(x)8=−250代入x=π−arcsin(132616−440002−1043953152+44000)+2π1−11cos(π−arcsin(132616−440002−1043953152+44000)+2π1)⋅22−11sin(π−arcsin(132616−440002−1043953152+44000)+2π1)⋅8=−250
整理后得231.24373…=−250
⇒假
检验 arcsin(132616440002−1043953152+44000)+2πn的解:真
arcsin(132616440002−1043953152+44000)+2πn
代入 n=1arcsin(132616440002−1043953152+44000)+2π1
对于 −11cos(x)22−11sin(x)8=−250代入x=arcsin(132616440002−1043953152+44000)+2π1−11cos(arcsin(132616440002−1043953152+44000)+2π1)⋅22−11sin(arcsin(132616440002−1043953152+44000)+2π1)⋅8=−250
整理后得−250=−250
⇒真
检验 π−arcsin(132616440002−1043953152+44000)+2πn的解:假
π−arcsin(132616440002−1043953152+44000)+2πn
代入 n=1π−arcsin(132616440002−1043953152+44000)+2π1
对于 −11cos(x)22−11sin(x)8=−250代入x=π−arcsin(132616440002−1043953152+44000)+2π1−11cos(π−arcsin(132616440002−1043953152+44000)+2π1)⋅22−11sin(π−arcsin(132616440002−1043953152+44000)+2π1)⋅8=−250
整理后得151.96794…=−250
⇒假
x=arcsin(132616−440002−1043953152+44000)+2πn,x=arcsin(132616440002−1043953152+44000)+2πn
以小数形式表示解x=0.10677…+2πn,x=0.59076…+2πn