Worked examples · Sheet 1

Elementary trigonometric equations

Theory used: Elementary equations Special values Unit circle, sine and cosine

Exercise 1

Solve the equation

sin x = \frac{2}{2} .

From the table of special values, an angle with sine $\frac{2}{2}$ is $α = \frac{π}{4}$ . The two families of solutions are

x = \frac{π}{4} + 2 nπ or x = π - \frac{π}{4} + 2 nπ = \frac{3 π}{4} + 2 nπ, n \in Z .

Solutions: $x = \frac{π}{4} + 2 nπ$ or $x = \frac{3 π}{4} + 2 nπ$ , with $n \in Z$ .

The equation is elementary, so one first looks for an angle $α$ with $sin α = \frac{2}{2}$ , and the table of special values gives $α = \frac{π}{4}$ . On the unit circle there are two points with ordinate $\frac{2}{2}$ , symmetric with respect to the vertical axis: the one of angle $\frac{π}{4}$ and the one of the supplementary angle $\frac{3 π}{4}$ . Since the sine is periodic with period $2 π$ , one adds $2 nπ$ to each angle and obtains the two families of solutions.

Exercise 2

Solve the equation

2 cos x + 1 = 0.

Isolate the cosine:

cos x = - \frac{1}{2} .

An angle with cosine $- \frac{1}{2}$ is $α = \frac{2 π}{3}$ , since $cos \frac{2 π}{3} = - cos \frac{π}{3} = - \frac{1}{2}$ . The solutions are

x = \frac{2 π}{3} + 2 nπ or x = - \frac{2 π}{3} + 2 nπ, n \in Z .

Solutions: $x = \pm \frac{2 π}{3} + 2 nπ$ , with $n \in Z$ .

The equation is not yet in elementary form, so one first isolates the trigonometric function by moving the constant term to the right-hand side and dividing by $2$ . The value $- \frac{1}{2}$ is obtained with related angles starting from $cos \frac{π}{3} = \frac{1}{2}$ , since the supplementary angle $\frac{2 π}{3}$ has the opposite cosine. On the unit circle there are two points with abscissa $- \frac{1}{2}$ , symmetric with respect to the horizontal axis, and they correspond to the angles $\frac{2 π}{3}$ and $- \frac{2 π}{3}$ ; adding whole turns gives the two families, which are written compactly with the double sign.

Exercise 3

Solve the equation

tan x = - 1.

The equation makes sense for $cos x \neq = 0$ , that is for $x \neq = \frac{π}{2} + nπ$ . An angle with tangent $- 1$ is $α = - \frac{π}{4}$ , since

tan (- \frac{π}{4}) = \frac{sin ( - \frac{π}{4} )}{cos ( - \frac{π}{4} )} = \frac{- \frac{2}{2}}{\frac{2}{2}} = - 1.

The tangent has period $π$ , therefore the solutions are

x = - \frac{π}{4} + nπ, n \in Z .

Solutions: $x = - \frac{π}{4} + nπ$ , with $n \in Z$ ; the family can also be written $x = \frac{3 π}{4} + nπ$ .

Before solving, one checks the domain, since the tangent is defined only where the cosine does not vanish; the solutions found do not fall at those points, given that $- \frac{π}{4} + nπ$ is never of the form $\frac{π}{2} + nπ$ . The angle $- \frac{π}{4}$ has tangent $- 1$ because sine and cosine have the same absolute value and opposite signs, therefore their ratio equals $- 1$ . Unlike sine and cosine, the tangent has period $π$ , hence the general solution is obtained by adding $nπ$ and a single family is enough. The form with $\frac{3 π}{4}$ describes the same set, given that $\frac{3 π}{4} = - \frac{π}{4} + π$ .