Real Trigonometric Equations

Definition: Trigonometric Equation

A trigonometric equation is an equation which contains variables in the arguments of real trigonometric functions.

Elementary Trigonometric Equations

Algorithm: Solving Equations of the Form $\sin x=c$

$$\sin x=c$$

Solutions:

• If $|c|>1$, then $x\in \varnothing$.
• If $c\in [-1;1]$, then

$$\begin{array}{rl}x& =\arcsin c+2k\pi \\ x& =-\arcsin c+(2k+1)\pi \end{array}k\in \mathbb{Z}$$

Algorithm: Solving Equations of the Form $\cos x=c$

$$\cos x=c$$

Solutions:

• If $|c|>1$, then $x\in \varnothing$.
• If $c\in [-1;1]$, then

$$x=\pm \arccos c+2k\pi k\in \mathbb{Z}$$

Algorithm: Solving Equations of the Form $\tan x=c$

$$\tan x=c$$

Requirements:

• $x\ne \frac{\pi }{2}+k\pi k\in \mathbb{Z}$

Solution:

$$x=\arctan c+k\pi k\in \mathbb{Z}$$

Algorithm: Solving Equations of the Form $\cot x=c$

$$\cot x=c$$

Requirements:

• $x\ne k\pi k\in \mathbb{Z}$

Solution:

$$x=\mathrm{arccot}+k\pi k\in \mathbb{Z}$$

Composed Trigonometric Equations

Algorithm: Solving Equations of the Form $\sin f(x)=\sin g(x)$

We are given a trigonometric equation of the form

$$\sin f(x)=\sin g(x)$$

Solution:

$$\begin{array}{rl}f(x)& =g(x)+2k\pi \\ f(x)& =-g(x)+(2k+1)\pi \end{array}k\in \mathbb{Z}$$

Algorithm: Solving Equations of the Form $\cos f(x)=\cos g(x)$

We are given a trigonometric equation of the form

$$\cos f(x)=\cos g(x)$$

Solution:

$$f(x)=\pm g(x)+2k\pi k\in \mathbb{Z}$$

Algorithm: Solving Equations of the Form $\tan f(x)=\tan g(x)$

We are given a trigonometric equation of the form

$$\tan f(x)=\tan g(x)$$

Requirements:

• $f(x),g(x)\ne \frac{\pi }{2}+k\pi k\in \mathbb{Z}$

Solution:

$$f(x)=g(x)+k\pi k\in \mathbb{Z}$$

Algorithm: Solving Equations of the Form $\cot f(x)=\cot g(x)$

We are given a trigonometric equation of the form

$$\cot f(x)=\cot g(x)$$

Requirements:

• $f(x),g(x)\ne k\pi k\in \mathbb{Z}$

Solution:

$$f(x)=g(x)+k\pi k\in \mathbb{Z}$$

Homogeneous Trigonometric Equations

Definition: Homogeneous Trigonometric Equation

A homogeneous trigonometric equation is a trigonometric equation of the form

$$a_0{\sin }^{n}x+a_1{\sin }^{n-1}x\cos x+\cdots +a_{n-1}\sin x{\cos }^{n-1}x+a_n{\cos }^{n}x=0,$$

where $a_k\in \mathbb{R}$.

Algorithm: Solving Homogeneous Trigonometric Equations (Tangent Substitution)

We are given the following homogeneous trigonometric equations.

$$a_0{\sin }^{n}x+a_1{\sin }^{n-1}x\cos x+\cdots +a_{n-1}\sin x{\cos }^{n-1}x+a_n{\cos }^{n}x=0$$

1. Check whether ${\cos }^n(x)=0$, i.e. $x=\pm \frac{\pi }{2}+2k\pi$ for $k\in \mathbb{Z}$, is a solution.
2. Divide by ${\cos }^n(x)$.

$$a_0\frac{{\sin }^{n}x}{{\cos }^{n}x}+a_1\frac{{\sin }^{n-1}x}{{\cos }^{n-1}x}+\cdots +a_{n-1}\frac{\sin x}{\cos x}+a_n=0$$ $$a_0{\tan }^{n}x+a_1{\tan }^{n-1}x+\cdots +a_{n-1}\tan x+a_n=0$$

3. Substitute $t=\tan x$ and solve the polynomial equation

$$a_0{t}^n+a_1{t}^{n-1}+\cdots +a_{n-1}t+a_n=0$$

4. For each solution $t^{\ast }$ to the equation in Step 3, solve the elementary trigonometric equation $\tan x=t^{\ast }$.

EXAMPLE

TODO

Algorithm: Solving Homogeneous Trigonometric Equations (Cotangent Substitution)

We are given the following homogeneous trigonometric equation.

$$a_0{\sin }^{n}x+a_1{\sin }^{n-1}x\cos x+\cdots +a_{n-1}\sin x{\cos }^{n-1}x+a_n{\cos }^{n}x=0$$

1. Check whether ${\sin }^n(x)=0$, i.e. $x=k\pi$ for $k\in \mathbb{Z}$, is a solution.
2. Divide by ${\sin }^n(x)$.

$$a_0+a_1\frac{\cos x}{\sin x}+\cdots +a_{n-1}\frac{{\cos }^{n-1}x}{{\sin }^{n-1}x}+a_n\frac{{\cos }^{n}x}{{\sin }^{n}x}=0$$ $$a_0+a_1\cot x+\cdots +a_{n-1}{\cot }^{n-1}x+a_n{\cot }^{n}x=0$$

3. Substitute $t=\cot x$ and solve the polynomial equation

$$a_0+a_{1}t+\cdots +a_{n-1}{t}^{n-1}+a_n{t}^n=0$$

4. For each solution $t^{\ast }$ to the equation in Step 3, solve the elementary trigonometric equation $\cot x=t^{\ast }$.

EXAMPLE

TODO

Other Trigonometric Equations

Algorithm: Solving Trigonometric Equations of the Form $a\sin x+b\cos x=c$

We are given a trigonometric equation of the following form.

$$a\sin x+b\cos x=c$$

1. Divide both sides by $\sqrt{a^2+b^2}$.

$$\frac{a}{\sqrt{a^2+b^2}}\sin x+\frac{b}{\sqrt{a^2+b^2}}\cos x=\frac{c}{\sqrt{a^2+b^2}}$$

2. Substitute $\cos \phi =\frac{a}{\sqrt{a^2+b^2}}$ and $\sin \phi =\frac{b}{\sqrt{a^2+b^2}}$

$$\cos \phi \sin x+\sin \phi \cos x=\frac{c}{\sqrt{a^2+b^2}}$$

• The reason we can do this is that ${\left(\frac{a}{\sqrt{a^2+b^2}}\right)}^2+{\left(\frac{b}{\sqrt{a^2+b^2}}\right)}^2=1$.

3. Use an identity to simplify the left-hand side.

$$\sin (x+\phi )=\frac{c}{\sqrt{a^2+b^2}}$$

• Requirements: $\frac{c}{\sqrt{a^2+b^2}}\in [-1;1]$

4. Solve the elementary trigonometric equation.

$$\begin{array}{rl}x+\phi & =\arcsin \frac{c}{\sqrt{a^2+b^2}}+2k\pi \\ x+\phi & =-\arcsin \frac{c}{\sqrt{a^2+b^2}}+(2k+1)\pi \end{array}$$

5. Rearrange to isolate $x$.

$$\begin{array}{rl}x& =\arcsin \frac{c}{\sqrt{a^2+b^2}}-\phi +2k\pi \\ x& =-\arcsin \frac{c}{\sqrt{a^2+b^2}}-\phi +(2k+1)\pi \end{array}$$

6.Substitute back for $\phi$ to obtain the final set of solutions.

• If you substitute back $\phi =\arccos \frac{a}{\sqrt{a^2+b^2}}$, you obtain

$$\begin{array}{rl}x& =\arcsin \frac{c}{\sqrt{a^2+b^2}}-\arccos \frac{a}{\sqrt{a^2+b^2}}+2k\pi \\ x& =-\arcsin \frac{c}{\sqrt{a^2+b^2}}-\arccos \frac{a}{\sqrt{a^2+b^2}}+(2k+1)\pi \end{array}$$

• If you susbtitute back $\phi =\arcsin \frac{b}{\sqrt{a^2+b^2}}$, you obtain

$$\begin{array}{rl}x& =\arcsin \frac{c}{\sqrt{a^2+b^2}}-\arcsin \frac{b}{\sqrt{a^2+b^2}}+2k\pi \\ x& =-\arcsin \frac{c}{\sqrt{a^2+b^2}}-\arcsin \frac{b}{\sqrt{a^2+b^2}}+(2k+1)\pi \end{array}$$

EXAMPLE

TODO