# Solving of Trigonometric Equatipns with a Method of Introducing New Variable

The essence of this method is in following: we try to transform given trigonometric equation into algebraic equation by introducing new variable.

**Example 1**. Solve equation `2cos^2(x)+14cos(x)=3sin^2(x)`.

Since `sin^2(x)=1-cos^2(x)` then equation can be rewritten as `2cos^2(x)+14cos(x)=3(1-cos^2(x))` or `5cos^2(x)+14cos(x)-3=0`.

Now, let `y=cos(x)` then we have that `5y^2+14y-3=0`. This quadratic equation has two roots: `y=1/5,y=-3`.

Thus, we obtained following set of equations: `cos(x)=1/5,cos(x)=-3`.

From first equation `x=+-arccos(1/5)+2pin,n in Z`; second equation doesn't have roots because `|cos(x)|<=1` for all x.

Therefore, initial equation has following solution: `x=+-arccos(1/5)+2pin,n in Z`.

Sometimes we can define rule that can make easier choice of new variable in trigonometric equations. Let R(sin(x),cos(x)) is rational function of sin(x) and cos(x), i.e. function that we obtain from functions sin(x), cos(x) and constants with the help of addition, subtraction, multiplication and division (for example `(sin^2(x)+0.5)/(sin(x)cos(x)-0.75)` is rational function, but `sin(x)-sqrt(cos(x))` is not because of the root).

Now, consider equation `R(sin(x),cos(x))=0`.

This equation can be transformed into algebraic by replacing `cos^2(x)` with `1-sin^2(x)`, only if equation has cos(x) raised to an even power.

Similarly this equation can be transformed into algebraic by replacing `sin^2(x)` with `1-cos^2(x)`, only if equation has sin(x) raised to an even power.

**Example 2**. Solve equation `cos^4(x)+3sin(x)-sin^4(x)-2=0`.

We can't replace `sin^2(x)` with `1-cos^2(x)` because equation contains sin(x) raised to 1 (odd power).

But we can replace `cos^2(x)` with `1-sin^2(x)` because equation contains cos(x) of even power.

So, `cos^2(x)=1-sin^2(x)` or `cos^4(x)=(1-sin^2(x))^2`.

Now, equation can be rewritten as `(1-sin^2(x))^2+3sin(x)-sin^4(x)-2=0`.

Let `y=sin(x)` then `(1-y^2)^2+3y-y^4-2=0` or `2y^2-3y+1=0`. This equation has two roots: `y=1,y=1/2`.

Therefore, we obtained set of equations: `sin(x)=1,sin(x)=1/2`.

From first equation `x=pi/2+pik,k in Z`, from second equation `x=(-1)^n pi/6+pin, n in Z`. All these roots are roots of the initial equation.