# Solution of trigonometrical equations

Introduction: Trigonometric equations involve trigonometric functions and are equations where the unknown variable is an angle. Solving these equations requires applying trigonometric identities, understanding periodic behavior, and employing algebraic techniques. Here are key concepts and steps for solving trigonometric equations:

### 1. Basic Concepts:

• Trigonometric Functions:
• Primary functions: sine ($\mathrm{sin}$), cosine ($\mathrm{cos}$), tangent ($\mathrm{tan}$), cotangent ($\mathrm{cot}$), secant ($\mathrm{sec}$), and cosecant ($\mathrm{csc}$).
• Periodicity:
• Sine and cosine have a principal period of $2\pi$, while tangent, cotangent, secant, and cosecant have a principal period of $\pi$.

### 2. Steps for Solving Trigonometric Equations:

1. Isolate the Trigonometric Function:

• Move all terms involving the trigonometric function to one side of the equation.
2. Apply Trigonometric Identities:

• Utilize trigonometric identities to simplify the equation. Common identities include ${\mathrm{sin}}^{2}\left(\theta \right)+{\mathrm{cos}}^{2}\left(\theta \right)=1$ and $\mathrm{tan}\left(\theta \right)=\frac{\mathrm{sin}\left(\theta \right)}{\mathrm{cos}\left(\theta \right)}$.
3. Use Algebraic Techniques:

• Employ algebraic techniques such as factoring, completing the square, or substitution.
4. Consider the Periodicity:

• Be aware of the periodic nature of trigonometric functions. If $x$ is a solution, then $x+2\pi n$ or $x+\pi n$ (where $n$ is an integer) is also a solution.
5. General Solutions:

• Express solutions in terms of general formulas that cover all possible solutions. For example, if $\mathrm{sin}\left(x\right)=0$, then $x=n\pi$ is the general solution.
6. Special Trigonometric Equations:

• Equations involving squares of trigonometric functions may be solved by factoring or applying the quadratic formula.
• Inverse Trigonometric Functions:
• Equations involving inverse trigonometric functions (${\mathrm{sin}}^{-1}\left(x\right)$, ${\mathrm{cos}}^{-1}\left(x\right)$, ${\mathrm{tan}}^{-1}\left(x\right)$ may require solving for the angle using the properties of inverse trigonometric functions.

Example: Solve the equation $\mathrm{cos}\left(2x\right)-\mathrm{sin}\left(x\right)=0$ for $0\le x\le 2\pi$.

Solution:

1. Isolate the Trigonometric Function: $\mathrm{cos}\left(2x\right)-\mathrm{sin}\left(x\right)=0$

2. Apply Trigonometric Identities:

• Use double-angle identity for cosine: $\mathrm{cos}\left(2x\right)=1-2{\mathrm{sin}}^{2}\left(x\right)$ $1-2{\mathrm{sin}}^{2}\left(x\right)-\mathrm{sin}\left(x\right)=0$
3. Combine Terms: $-2{\mathrm{sin}}^{2}\left(x\right)-\mathrm{sin}\left(x\right)+1=0$

• Recognize the quadratic form ($a{x}^{2}+bx+c=0$) where $a=-2$, $b=-1$, and $c=1$. $-2{\mathrm{sin}}^{2}\left(x\right)+\mathrm{sin}\left(x\right)-1=0$

• Solve the quadratic equation using factoring, completing the square, or the quadratic formula. Here, we'll factor: $\left(\mathrm{sin}\left(x\right)-1\right)\left(2\mathrm{sin}\left(x\right)+1\right)=0$

This gives two possible equations: $\mathrm{sin}\left(x\right)-1=0\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}2\mathrm{sin}\left(x\right)+1=0$

6. Solve Each Equation:

• For $\mathrm{sin}\left(x\right)-1=0$: $\mathrm{sin}\left(x\right)=1\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}x=\frac{\pi }{2}+2n\pi$

• For $2\mathrm{sin}\left(x\right)+1=0$:

7. Combine Solutions:

• Combine all solutions: $x=\frac{\pi }{2}+2n\pi ,\phantom{\rule{1em}{0ex}}x=\frac{7\pi }{6}+2n\pi ,\phantom{\rule{1em}{0ex}}x=\frac{11\pi }{6}+2n\pi$ where $n$ is an integer.

Example: Solve the Trigonometric Equation

${\mathrm{tan}}^{2}\left(x\right)+3\mathrm{tan}\left(x\right)+2=0$

Solution:

• The given equation is quadratic in $\mathrm{tan}\left(x\right)$, so we can solve it using standard quadratic equation techniques.
• The quadratic formula is given by: $\mathrm{tan}\left(x\right)=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
• For our equation ${\mathrm{tan}}^{2}\left(x\right)+3\mathrm{tan}\left(x\right)+2=0$, the coefficients are $a=1$, $b=3$, and $c=2$.
• $\mathrm{tan}\left(x\right)=\frac{-3±\sqrt{{3}^{2}-4\left(1\right)\left(2\right)}}{2\left(1\right)}$
• $\mathrm{tan}\left(x\right)=\frac{-3±\sqrt{1}}{2}$$\mathrm{tan}\left(x\right)=\frac{-3±1}{2}$
• Solution 1: $\mathrm{tan}\left(x\right)=\frac{-3+1}{2}=-1$
• This implies $x={\mathrm{tan}}^{-1}\left(-1\right)=-\frac{\pi }{4}+n\pi$ where $n$ is an integer.
• Solution 2: $\mathrm{tan}\left(x\right)=\frac{-3-1}{2}=-2$
• This has no real solution since the range of the tangent function is $\left(-\mathrm{\infty },\mathrm{\infty }\right)$