解答
cos2(θ)+9.81(1)102cos(θ)−1=0
解答
θ=1.47346…+2πn,θ=2π−1.47346…+2πn
+1
度数
θ=84.42355…∘+360∘n,θ=275.57644…∘+360∘n求解步骤
cos2(θ)+9.81(1)102cos(θ)−1=0
用替代法求解
cos2(θ)+9.81⋅1102cos(θ)−1=0
令:cos(θ)=uu2+9.81⋅1102u−1=0
u2+9.81⋅1102u−1=0:u=19.62−100+10384.9444,u=19.62−100−10384.9444
u2+9.81⋅1102u−1=0
在两边乘以 9.81
u2+9.81⋅1102u−1=0
在两边乘以 9.81u2⋅9.81+9.81⋅1102u⋅9.81−1⋅9.81=0⋅9.81
化简9.81u2+100u−9.81=0
9.81u2+100u−9.81=0
使用求根公式求解
9.81u2+100u−9.81=0
二次方程求根公式:
若 a=9.81,b=100,c=−9.81u1,2=2⋅9.81−100±1002−4⋅9.81(−9.81)
u1,2=2⋅9.81−100±1002−4⋅9.81(−9.81)
1002−4⋅9.81(−9.81)=10384.9444
1002−4⋅9.81(−9.81)
使用法则 −(−a)=a=1002+4⋅9.81⋅9.81
数字相乘:4⋅9.81⋅9.81=384.9444=1002+384.9444
1002=10000=10000+384.9444
数字相加:10000+384.9444=10384.9444=10384.9444
u1,2=2⋅9.81−100±10384.9444
将解分隔开u1=2⋅9.81−100+10384.9444,u2=2⋅9.81−100−10384.9444
u=2⋅9.81−100+10384.9444:19.62−100+10384.9444
2⋅9.81−100+10384.9444
数字相乘:2⋅9.81=19.62=19.62−100+10384.9444
u=2⋅9.81−100−10384.9444:19.62−100−10384.9444
2⋅9.81−100−10384.9444
数字相乘:2⋅9.81=19.62=19.62−100−10384.9444
二次方程组的解是:u=19.62−100+10384.9444,u=19.62−100−10384.9444
u=cos(θ)代回cos(θ)=19.62−100+10384.9444,cos(θ)=19.62−100−10384.9444
cos(θ)=19.62−100+10384.9444,cos(θ)=19.62−100−10384.9444
cos(θ)=19.62−100+10384.9444:θ=arccos(19.62−100+10384.9444)+2πn,θ=2π−arccos(19.62−100+10384.9444)+2πn
cos(θ)=19.62−100+10384.9444
使用反三角函数性质
cos(θ)=19.62−100+10384.9444
cos(θ)=19.62−100+10384.9444的通解cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πnθ=arccos(19.62−100+10384.9444)+2πn,θ=2π−arccos(19.62−100+10384.9444)+2πn
θ=arccos(19.62−100+10384.9444)+2πn,θ=2π−arccos(19.62−100+10384.9444)+2πn
cos(θ)=19.62−100−10384.9444:无解
cos(θ)=19.62−100−10384.9444
−1≤cos(x)≤1无解
合并所有解θ=arccos(19.62−100+10384.9444)+2πn,θ=2π−arccos(19.62−100+10384.9444)+2πn
以小数形式表示解θ=1.47346…+2πn,θ=2π−1.47346…+2πn