Systems of Trigonometric Equations

Whe we solve systems of trigonometric equations, we use trigonometric formulas and standard techniques for solving systems of equations.

Example. Solve the following system of equations: {(sin(x)+cos(y)=1.5),(sin^2(x)+cos^2(y)=1.25):}.

Let's make substitution u=sin(x),v=cos(x), then equation can be rewritten as {(u+v=1.5),(u^2+v^2=1.25):}.

Express v from first equation: v=1.5-u. Now, plug this expression into second equation: u^2+(1.5-u)^2=1.25 or 2u^2-3u+1=0. This equation has two roots: u_1=1,u_2=0.5.

Now, use equation v=1.5-u.

1. v_1=1.5-u_1=1.5-1=0.5;
2. v_2=1.5-u_2=1.5-0.5=1.

Thus, we obtained two pairs of values: u_1=1,v_1=0.5 and u_2=0.5,v_2=1.

Now, if we return to old variables, we will obtain set of two systems: {(sin(x)=1),(cos(x)=0.5):},{(sin(x)=0.5),(cos(x)=1):}.

Find solutions of first system: from equation sin(x)=1 we have that x=pi/2+2pik,k in Z; from equation cos(y)=0.5 we have that y=+-pi/3+2pin,n in Z. Therefore, this system has solutions of the form x=pi/2+2pik,\ y=+-pi/3+2pin;\ k,n in Z.

Now find solutions of first system: from equation sin(x)=0.5 we have that x=(-1)^m pi/6+pim,m in Z; from equation cos(y)=1 we have that y=2pil,l in Z. Therefore, this system has solutions of the form x=(-1)^m pi/6+pim,\ y=2pil;\ m,l in Z.

Thus, initial system has solutions of the form x=pi/2+2pik,\ y=+-pi/3+2pin;\ k,n in Z and x=(-1)^m pi/6+pim,\ y=2pil;\ m,l in Z.