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`.