# 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.