matlab 解个方程

你是不是写错了啊？
f1=sym('pi/2*(cos(a))^2+x*sin(2*a)=0.2');
是不是改成f1=sym('pi/2*(cos(a))^2+x*sin(2*a)-0.2');
或者直接f1=pi/2*(cos(a))^2+x*sin(2*a)-0.2，同时要定义a为syms a；
我改成syms a;x=0.7;f1 =pi/2*(cos(a))^2+x*sin(2*a)-0.2;solve(f1,a)，运行是成功的
结果为：c=atan(1/2*(5/14*2^(1/2)*((5*pi+49+7*(10*pi+45)^(1/2))/(25*pi^2+196))^(1/2)*pi+7*2^(1/2)*((5*pi+49+7*(10*pi+45)^(1/2))/(25*pi^2+196))^(1/2)-14*2^(1/2)*((5*pi+49+7*(10*pi+45)^(1/2))/(25*pi^2+196))^(3/2)-25/14*2^(1/2)*((5*pi+49+7*(10*pi+45)^(1/2))/(25*pi^2+196))^(3/2)*pi^2)*2^(1/2)/((5*pi+49+7*(10*pi+45)^(1/2))/(25*pi^2+196))^(1/2))
-atan(1/2*(-5/14*2^(1/2)*((5*pi+49+7*(10*pi+45)^(1/2))/(25*pi^2+196))^(1/2)*pi-7*2^(1/2)*((5*pi+49+7*(10*pi+45)^(1/2))/(25*pi^2+196))^(1/2)+14*2^(1/2)*((5*pi+49+7*(10*pi+45)^(1/2))/(25*pi^2+196))^(3/2)+25/14*2^(1/2)*((5*pi+49+7*(10*pi+45)^(1/2))/(25*pi^2+196))^(3/2)*pi^2)*2^(1/2)/((5*pi+49+7*(10*pi+45)^(1/2))/(25*pi^2+196))^(1/2))+pi
atan(1/2*(5/14*2^(1/2)*((5*pi+49-7*(10*pi+45)^(1/2))/(25*pi^2+196))^(1/2)*pi+7*2^(1/2)*((5*pi+49-7*(10*pi+45)^(1/2))/(25*pi^2+196))^(1/2)-14*2^(1/2)*((5*pi+49-7*(10*pi+45)^(1/2))/(25*pi^2+196))^(3/2)-25/14*2^(1/2)*((5*pi+49-7*(10*pi+45)^(1/2))/(25*pi^2+196))^(3/2)*pi^2)*2^(1/2)/((5*pi+49-7*(10*pi+45)^(1/2))/(25*pi^2+196))^(1/2))
-atan(1/2*(-5/14*2^(1/2)*((5*pi+49-7*(10*pi+45)^(1/2))/(25*pi^2+196))^(1/2)*pi-7*2^(1/2)*((5*pi+49-7*(10*pi+45)^(1/2))/(25*pi^2+196))^(1/2)+14*2^(1/2)*((5*pi+49-7*(10*pi+45)^(1/2))/(25*pi^2+196))^(3/2)+25/14*2^(1/2)*((5*pi+49-7*(10*pi+45)^(1/2))/(25*pi^2+196))^(3/2)*pi^2)*2^(1/2)/((5*pi+49-7*(10*pi+45)^(1/2))/(25*pi^2+196))^(1/2))-pi