Universal Substitution for Rational Trigonometric Equations

It is know that sin(x) and cos(x) can be rationally expressed through `tan(x/2)`, namely:

`sin(x)=(2tan(x/2))/(1+tan^2(x/2)),\ cos(x)=(1-tan^2(x/2))/(1+tan^2(x/2))`, where `x!=pi+2pin, n in Z` (i.e. where `tan(x/2)` is not defined).

Substitution `u=tan(x/2)` is called universal. It can be used to solve any equation of the form R(sin(x),cos(x))=0 (i.e. any rational equation of sin(x) and cos(x)).

Since we can apply universal substitution only when `x!=pi+2pin,n in Z` then we need to check whether `x=pi+2pin,n in Z` are roots of given equation or not.

Example 1. Solve `3sin(x)+4cos(x)=5`.

Using above formulas and making substitution `u=tan(x/2)` we obtain that `3*(2u)/(1+u^2)+4*(1-u^2)/(1+u^2)=5` or `6u+4(1-u^2)=5(1+u^2)`. This leads to the equation `9u^2-6u+1=0` or `(3u-1)^2=0`. This equation has one root `u=1/3`.

Thus, we obtain equation `tan(x/2)=1/3`. So, `x/2=arctan(1/3)+pin,n in Z`, i.e. `x=2arctan(1/3)+2pin,n in Z`.

Now, we need to check whether `x=pi+2pik,k in Z` is root of the equation.

If `x=pi+2pik, k in Z`, then `sin(x)=0` and `cos(x)=-1`. Plugging these values into initial equation gives `3*0+4*(-1)=-4!=5`. Thus, `x=pi+2pik,k in Z` is not root of the given equation.

So, solutions of the initial equation are `x=2arctan(1/3)+2pin, n in Z`.

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

Let's use universal substitution. Here `sin(2x)=(2tan((2x)/2))/(1+tan^2((2x)/2))` or `sin(2x)=(2tan(x))/(1+tan^2(x))` and `cos(2x)=(1-tan^2((2x)/2))/(1+tan^2((2x)/2))` or `cos(2x)=(1-tan^2(x))/(1+tan^2(x))`.

So, equation can be rewritten as `3*(2tan(x))/(1+tan^2(x))+(1-tan^2(x))/(1+tan^2(x))+1=0`.

Now, let `u=tan(x)` then `3 (2u)/(1+u^2)+(1-u^2)/(1+u^2)+1=0`. This can be rewritten as `6u+(1-u^2)+(1+u^2)=0` or `6u=-2`. So, `u=-1/3`.

Thus, `tan(x)=-1/3` or `x=arctan(-1/3)+pin, n in Z`.

Now, we need to check whether `2x=pi+2pik,k in Z` is root of the equation.

If `2x=pi+2pik, k in Z`, then `sin(2x)=0` and `cos(2x)=-1`. Plugging these values into initial equation gives `3*0-1+1=0`. Thus, `2x=pi+2pik,k in Z` or `x=pi/2+pik,k in Z` is root of the given equation.

Therefore, given equation has following solutions: `x=arctan(-1/3)+pin;\ x=pi/2+pik;\ k,n in Z`.