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`.