解答
cos(20∘)+14sin2(θ)=10
解答
θ=0.93477…+360∘n,θ=180∘−0.93477…+360∘n,θ=−0.93477…+360∘n,θ=180∘+0.93477…+360∘n
+1
弧度
θ=0.93477…+2πn,θ=π−0.93477…+2πn,θ=−0.93477…+2πn,θ=π+0.93477…+2πn求解步骤
cos(20∘)+14sin2(θ)=10
用替代法求解
cos(20∘)+14sin2(θ)=10
令:sin(θ)=ucos(20∘)+14u2=10
cos(20∘)+14u2=10:u=1410−cos(20∘),u=−1410−cos(20∘)
cos(20∘)+14u2=10
将 cos(20∘)到右边
cos(20∘)+14u2=10
两边减去 cos(20∘)cos(20∘)+14u2−cos(20∘)=10−cos(20∘)
化简14u2=10−cos(20∘)
14u2=10−cos(20∘)
两边除以 14
14u2=10−cos(20∘)
两边除以 141414u2=1410−14cos(20∘)
化简
1414u2=1410−14cos(20∘)
化简 1414u2:u2
1414u2
数字相除:1414=1=u2
化简 1410−14cos(20∘):1410−cos(20∘)
1410−14cos(20∘)
使用法则 ca±cb=ca±b=1410−cos(20∘)
u2=1410−cos(20∘)
u2=1410−cos(20∘)
u2=1410−cos(20∘)
对于 x2=f(a) 解为 x=f(a),−f(a)
u=1410−cos(20∘),u=−1410−cos(20∘)
u=sin(θ)代回sin(θ)=1410−cos(20∘),sin(θ)=−1410−cos(20∘)
sin(θ)=1410−cos(20∘),sin(θ)=−1410−cos(20∘)
sin(θ)=1410−cos(20∘):θ=arcsin(1410−cos(20∘))+360∘n,θ=180∘−arcsin(1410−cos(20∘))+360∘n
sin(θ)=1410−cos(20∘)
使用反三角函数性质
sin(θ)=1410−cos(20∘)
sin(θ)=1410−cos(20∘)的通解sin(x)=a⇒x=arcsin(a)+360∘n,x=180∘−arcsin(a)+360∘nθ=arcsin(1410−cos(20∘))+360∘n,θ=180∘−arcsin(1410−cos(20∘))+360∘n
θ=arcsin(1410−cos(20∘))+360∘n,θ=180∘−arcsin(1410−cos(20∘))+360∘n
sin(θ)=−1410−cos(20∘):θ=arcsin(−1410−cos(20∘))+360∘n,θ=180∘+arcsin(1410−cos(20∘))+360∘n
sin(θ)=−1410−cos(20∘)
使用反三角函数性质
sin(θ)=−1410−cos(20∘)
sin(θ)=−1410−cos(20∘)的通解sin(x)=−a⇒x=arcsin(−a)+360∘n,x=180∘+arcsin(a)+360∘nθ=arcsin(−1410−cos(20∘))+360∘n,θ=180∘+arcsin(1410−cos(20∘))+360∘n
θ=arcsin(−1410−cos(20∘))+360∘n,θ=180∘+arcsin(1410−cos(20∘))+360∘n
合并所有解θ=arcsin(1410−cos(20∘))+360∘n,θ=180∘−arcsin(1410−cos(20∘))+360∘n,θ=arcsin(−1410−cos(20∘))+360∘n,θ=180∘+arcsin(1410−cos(20∘))+360∘n
以小数形式表示解θ=0.93477…+360∘n,θ=180∘−0.93477…+360∘n,θ=−0.93477…+360∘n,θ=180∘+0.93477…+360∘n