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Solving Trigonometric Equations using Algebraic Methods

An equation that contains trigonometric functions is called a trigonometric equation .

Example:

$\begin{array}{l} \sin^{2} x + \cos^{2} x = 1 \\ 2 \sin x - 1 = 0 \\ \tan^{2} 2 x - 1 = 0 \end{array}$

Solving Trigonometric Equations

To solve a trigonometric equation, we use the rules of algebra to isolate the trigonometric function on one side of the equal sign. Then we use our knowledge of the values of the trigonometric functions to solve for the variable.

When you solve a trigonometric equation that involves only one trigonometric expression, begin by isolating the expression.

When trigonometric functions cannot be combined on one side of an equation, try to factor the equation and then apply zero product property to solve the equation. If the equation has quadratic form, first factor if possible. If not possible, apply the quadratic formula to solve the equation.

Example :

Solve $2 \sin x - 1 = 0$ .

To solve the equation, we begin by rewriting it so that sin $x$ is isolated on the left side. So, first add $1$ to each side and then divide each side by $2$ .

$\begin{array}{l} 2 \sin x = 1 \\ \sin x = \frac{1}{2} \end{array}$

Since $\sin x$ has a period of $2 π$ , first we find all the solutions in the interval $[0, 2 π]$ .

The solutions are $x = \frac{π}{6}$ and $x = \frac{5 π}{6}$ .

The solutions on the interval $(- \infty, \infty)$ are then found by adding integer multiples of $2 π$ . Adding $2 n π$ to each of the solutions, we obtain the general solution of the given equation. Therefore, the general form of the solutions is $x = \frac{π}{6} + 2 n π$ and $x = \frac{5 π}{6} + 2 n π$ where $n$ is any integer.