# Homogeneous Trigonometric Equations

In many cases trigonometric equation of the form R(sin(x),cos(x)=0) can be transformed into algebraic with respect to tan(x). Main property of such equations is that they don't change if we replace sin(x) with -sin(x) and cos(x) with -cos(x).

Examples of such equations are homogeneous equations:

1. asin(x)+bcos(x)=0; (homogeneous equation of first degree). Indeed if we replace sin(x) with -sin(x) and cos(x) with -cos(x) then we will obtain that -asin(x)-bcos(x)=0 or asin(x)+bcos(x)=0.
2. a\ sin^2(x)+b\ sin(x)cos(x)+c\ cos^2(x)=0 (homogeneous equation of second degree).

Consider case when a!=0.

If we divide both sides of equation asin(x)+bcos(x)=0 by cos(x) then we will obtain that atan(x)+b=0. This equation is algebraic with respect to tan(x).

If we divide both sides of equation a\ sin^2(x)+b\ sin(x)cos(x)+c\ cos^2(x)=0 by cos^2(x) then we will obtain that a\ tan^2(x)+b\ tan(x)+c=0=0. This equation is algebraic with respect to tan(x).

Note, that when a!=0, roots of the equation cos(x)=0 are not roots of homogeneous equation. Therefore, when we divide by cos(x) (or cos^2(x)) we don't lose any roots (in case a!=0).

Example 1. Solve equation sin^2(x)+2sin(x)cos(x)-3cos^2(x)=0.

Dividing both sides by cos^2(x) gives tan^2(x)+2tan(x)-3=0.

Let u=tan(x) then u^2+2u-3=0. This equation has two roots: u=-3, u=1.

Therefore, we have set of equations: tan(x)=-3,tan(x)=1.

From first equation x=arctan(-3)+pik;\ x=pi/4+pin;\ k,n in Z. All these values are roots of initial equation.

Example 2. Solve equation 5sin^2(x)+sqrt(3)sin(x)cos(x)+6cos^2(x)=5.

Since sin^2(x)+cos^2(x)=1 then we can rewrite equation as 5sin^2(x)+sqrt(3)sin(x)cos(x)+6cos^2(x)=5(sin^2(x)+cos^2(x)) or

 sqrt(3)sin(x)cos(x)+cos^2(x)=0 .

In this equation there is no member of the form a\ sin^2(x), i.e. a=0. In this case we can't divide by cos^2(x), because roots of the equation cos(x)=0 are also roots of the initial equation, i.e. we will lose roots.

Instead, we factor left side: cos(x)(sqrt(3)sin(x)+cos(x))=0.

So, either cos(x)=0 or sqrt(3)sin(x)+cos(x)=0.

From first equation x=pi/2+pik, k in Z.

To solve second equation we divide both sides by cos(x) (we can do it and we will not lose roots, because we've already considered case when cos(x)=0): sqrt(3)tan(x)+1=0 or tan(x)=-1/(sqrt(3)). From this we have that x=arctan(-1/(sqrt(3)))+pin, n in Z, i.e. x=-pi/6+pik,k in Z.

Thus, solutions of the initial equation are x=pi/2+pik;\ x=-pi/6+pin;\ k,n in Z.