Trigonometry/Solving Trigonometric Equations

Trigonometric equations are equations including trigonometric functions. If they have only such functions and constants, then the solution involves finding an unknown which is an argument to a trigonometric function.

sin(x) = n
The equation $$\sin(x)=n$$ has solutions only when $$n$$ is within the interval $$[-1,1]$$. If $$n$$ is within this interval, then we first find an $$\alpha$$ such that:
 * $$\alpha=\arcsin(n)$$

The solutions are then:
 * $$x=\alpha+2k\pi$$
 * $$x=\pi-\alpha+2k\pi$$

Where $$k$$ is an integer.

In the cases when $$n$$ equals 1, 0 or -1 these solutions have simpler forms which are summarized in the table on the right.

For example, to solve:
 * $$\sin\bigl(\tfrac{x}{2}\bigr)=\frac{\sqrt3}{2}$$

First find $$\alpha$$ :
 * $$\alpha=\arcsin\bigl(\tfrac{\sqrt3}{2}\bigr)=\frac{\pi}{3}$$

Then substitute in the formulae above:
 * $$\frac{x}{2}=\frac{\pi}{3}+2k\pi$$


 * $$\frac{x}{2}=\pi-\frac{\pi}{3}+2k\pi$$

Solving these linear equations for $$x$$ gives the final answer:
 * $$x=\frac{2\pi}{3}(1+6k)$$


 * $$x=\frac{4\pi}{3}(1+3k)$$

Where $$k$$ is an integer.

cos(x) = n
Like the sine equation, an equation of the form $$\cos(x)=n$$ only has solutions when n is in the interval $$[-1,1]$$. To solve such an equation we first find one angle $$\alpha$$ such that:
 * $$\alpha=\arccos(n)$$

Then the solutions for $$x$$ are:
 * $$x=\pm\alpha+ 2k\pi$$

Where $$k$$ is an integer.

Simpler cases with $$n$$ equal to 1, 0 or -1 are summarized in the table on the right.

tan(x) = n
An equation of the form $$\tan(x)=n$$ has solutions for any real $$n$$. To find them we must first find an angle $$\alpha$$ such that:
 * $$\alpha=\arctan(n)$$

After finding $$\alpha$$, the solutions for $$x$$ are:
 * $$x=\alpha+k\pi$$

When $$n$$ equals 1, 0 or -1 the solutions have simpler forms which are shown in the table on the right.

cot(x) = n
The equation $$\cot(x)=n$$ has solutions for any real $$n$$. To find them we must first find an angle $$\alpha$$ such that:
 * $$\alpha=\arccot(n)$$

After finding $$\alpha$$, the solutions for $$x$$ are:
 * $$x=\alpha+k\pi$$

When $$n$$ equals 1, 0 or -1 the solutions have simpler forms which are shown in the table on the right.

csc(x) = n and sec(x) = n
The trigonometric equations $$\csc(x)=n$$ and $$\sec(x)=n$$ can be solved by transforming them to other basic equations:
 * $$\csc(x)=n\ \Leftrightarrow\ \frac{1}{\sin(x)}=n\ \Leftrightarrow\ \sin(x)=\frac{1}{n}$$
 * $$\sec(x)=n\ \Leftrightarrow\ \frac{1}{\cos(x)}=n\ \Leftrightarrow\ \cos(x)=\frac{1}{n}$$

Further examples
Generally, to solve trigonometric equations we must first transform them to a basic trigonometric equation using the trigonometric identities. This sections lists some common examples.

a sin(x)+b cos(x) = c
To solve this equation we will use the identity:
 * $$a\sin(x)+b\cos(x)=\sqrt{a^2+b^2}\sin(x+\alpha)$$


 * $$\alpha=\begin{cases}\arctan\bigl(\frac{b}{a}\bigr), & \mbox{if } a>0 \\ \pi+\arctan\bigl(\frac{b}{a}\bigr), & \mbox{if } a<0 \end{cases}$$

The equation becomes:
 * $$\sqrt{a^2+b^2}\sin(x+\alpha)=c$$


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

This equation is of the form $$\sin(x)=n$$ and can be solved with the formulae given above.

For example we will solve:
 * $$\sin(3x)-\sqrt3\cos(3x)=-\sqrt3$$

In this case we have:
 * $$a=1,b=-\sqrt3$$
 * $$\sqrt{a^2+b^2}=\sqrt{1^2+\Big(-\sqrt3\Big)^2}=2$$
 * $$\alpha=\arctan\Big(-\sqrt3\Big)=-\frac{\pi}{3}$$

Apply the identity:
 * $$2\sin\left(3x-\frac{\pi}{3}\right)=-\sqrt3$$
 * $$\sin\left(3x-\frac{\pi}{3}\right)=-\frac{\sqrt3}{2}$$

So using the formulae for $$\sin(x)=n$$ the solutions to the equation are:
 * $$3x-\frac{\pi}{3}=-\frac{\pi}{3}+2k\pi\ \Leftrightarrow\ x=\frac{2k\pi}{3}$$


 * $$3x-\frac{\pi}{3}=\pi+\frac{\pi}{3}+2k\pi\ \Leftrightarrow\ x=\frac{\pi}{9}(6k+5)$$

Where $$k$$ is an integer.

Matemática elementar/Trigonometria/Equações e inequações envolvendo funções trigonométricas