解答
sin2(x)+cos2(52)=1
解答
x=0.4+2πn,x=π−0.4+2πn,x=−0.4+2πn,x=π+0.4+2πn
+1
度数
x=22.91831…∘+360∘n,x=157.08168…∘+360∘n,x=−22.91831…∘+360∘n,x=202.91831…∘+360∘n求解步骤
sin2(x)+cos2(52)=1
用替代法求解
sin2(x)+cos2(52)=1
令:sin(x)=uu2+cos2(52)=1
u2+cos2(52)=1:u=1−cos2(52),u=−1−cos2(52)
u2+cos2(52)=1
将 cos2(52)到右边
u2+cos2(52)=1
两边减去 cos2(52)u2+cos2(52)−cos2(52)=1−cos2(52)
化简u2=1−cos2(52)
u2=1−cos2(52)
对于 x2=f(a) 解为 x=f(a),−f(a)
u=1−cos2(52),u=−1−cos2(52)
u=sin(x)代回sin(x)=1−cos2(52),sin(x)=−1−cos2(52)
sin(x)=1−cos2(52),sin(x)=−1−cos2(52)
sin(x)=1−cos2(52):x=arcsin(1−cos2(52))+2πn,x=π−arcsin(1−cos2(52))+2πn
sin(x)=1−cos2(52)
使用反三角函数性质
sin(x)=1−cos2(52)
sin(x)=1−cos2(52)的通解sin(x)=a⇒x=arcsin(a)+2πn,x=π−arcsin(a)+2πnx=arcsin(1−cos2(52))+2πn,x=π−arcsin(1−cos2(52))+2πn
x=arcsin(1−cos2(52))+2πn,x=π−arcsin(1−cos2(52))+2πn
sin(x)=−1−cos2(52):x=arcsin(−1−cos2(52))+2πn,x=π+arcsin(1−cos2(52))+2πn
sin(x)=−1−cos2(52)
使用反三角函数性质
sin(x)=−1−cos2(52)
sin(x)=−1−cos2(52)的通解sin(x)=−a⇒x=arcsin(−a)+2πn,x=π+arcsin(a)+2πnx=arcsin(−1−cos2(52))+2πn,x=π+arcsin(1−cos2(52))+2πn
x=arcsin(−1−cos2(52))+2πn,x=π+arcsin(1−cos2(52))+2πn
合并所有解x=arcsin(1−cos2(52))+2πn,x=π−arcsin(1−cos2(52))+2πn,x=arcsin(−1−cos2(52))+2πn,x=π+arcsin(1−cos2(52))+2πn
以小数形式表示解x=0.4+2πn,x=π−0.4+2πn,x=−0.4+2πn,x=π+0.4+2πn