Solving of Trigonometric Equations by Factoring

If trigonometric equation has form f(x)=0, and left side f(x) can be factored, then we need to equal to zero all factors. As result, we will obtain set of equations, and roots of each of them will be roots of the initial equation if they belong to the domain of each factor in the left part of equation.

Example 1. Solve equation `cos(x)(tan(x)-1)=0`.

We obtain set of equations: `cos(x)=0, tan(x)-1=0`.

From first equation `x=pi/2+pik,k in Z`, from second equation `x=pi/4+pin, n in Z`. However, values `x=pi/2+pik` are extraneous roots of the initial equation, because for this values `tan(x)` is not defined.

Therefore, solution of the initial equation has form `x=pi/4+pin,n in Z`.

Example 2. Solve equation `sin(5x)+sin(x)+2sin^2(x)=1`.

Convert sum `sin(5x)+sin(x)` into product: `sin(5x)+sin(x)=2sin((5x+x)/2)cos((5x-x)/2)=2sin(3x)cos(2x)`. Also, `2sin^2(x)=1-cos(2x)`.

Now equation can be rewritten as `2sin(3x)cos(2x)+1-cos(2x)=1` or `cos(2x)(2sin(3x)-1)=0`.

Therefore, we obtained set of equations: `cos(2x)=0,\ 2sin(3x)-1=0`.

From first equation `2x=pi/2+pik,k in Z`, i.e. `x=pi/4+(pik)/2,k in Z`.

From second equation `sin(3x)=1/2` or `3x=(-1)^n arcsin(1/2)+pin,n in Z`. Since `arcsin(1/2)=pi/6` then `3x=(-1)^n pi/6+pin,n in Z` or `x=(-1)^n pi/18+(pin)/3,n in Z`.

Therefore, asnwer is `x=pi/4+(pik)/2;\ x=(-1)^n (pi)/(18)+(pin)/3;\ k,n in Z`.