Worked examples · Sheet 2

Equations reducible to elementary ones

Theory used: Equality of two sines or of two cosines Related angles Elementary equations

Exercise 1

Solve the equation

sin 3 x = sin x .

Two angles have the same sine when they coincide up to full turns or are supplementary up to full turns:

3 x = x + 2 nπ or 3 x = π - x + 2 nπ .

From the first:

2 x = 2 nπ, x = nπ .

From the second:

4 x = π + 2 nπ, x = \frac{π}{4} + \frac{nπ}{2} .

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

The equation sets two sines equal, so one applies the two alternatives from the theory: the arguments coincide up to an integer number of turns, or they are supplementary up to full turns. Each alternative is a linear equation in $x$ and is solved by isolating $x$ ; the final division also involves the term $2 nπ$ , hence in the second family the step between two consecutive solutions reduces to $\frac{π}{2}$ . A check on one value: for $x = \frac{π}{4}$ one has $sin \frac{3 π}{4} = \frac{2}{2} = sin \frac{π}{4}$ , therefore the equality is satisfied.

Exercise 2

Solve the equation

cos (2 x - \frac{π}{3}) = cos x .

Two angles have the same cosine when they coincide or are opposite, up to full turns:

2 x - \frac{π}{3} = x + 2 nπ or 2 x - \frac{π}{3} = - x + 2 nπ .

From the first:

x = \frac{π}{3} + 2 nπ .

From the second:

3 x = \frac{π}{3} + 2 nπ, x = \frac{π}{9} + \frac{2 nπ}{3} .

Solutions: $x = \frac{π}{3} + 2 nπ$ or $x = \frac{π}{9} + \frac{2 nπ}{3}$ , with $n \in Z$ .

For the cosine the two alternatives are equal or opposite arguments, always up to full turns, because on the unit circle the points with the same abscissa are symmetric with respect to the horizontal axis. The first alternative gives $x$ immediately by moving $x$ to the left and $\frac{π}{3}$ to the right; the second moves $- x$ to the left, so one divides the whole right-hand side by $3$ , including the term with $n$ . A check on the first family: for $x = \frac{π}{3}$ the argument on the left equals $\frac{2 π}{3} - \frac{π}{3} = \frac{π}{3}$ , equal to the argument on the right, therefore the equality is satisfied.

Exercise 3

Solve the equation

sin 2 x = cos x .

With related angles the cosine is written as a sine:

cos x = sin (\frac{π}{2} - x),

therefore the equation becomes $sin 2 x = sin (\frac{π}{2} - x)$ and the alternatives are

2 x = \frac{π}{2} - x + 2 nπ or 2 x = π - (\frac{π}{2} - x) + 2 nπ .

From the first:

3 x = \frac{π}{2} + 2 nπ, x = \frac{π}{6} + \frac{2 nπ}{3} .

From the second:

2 x = \frac{π}{2} + x + 2 nπ, x = \frac{π}{2} + 2 nπ .

Solutions: $x = \frac{π}{6} + \frac{2 nπ}{3}$ or $x = \frac{π}{2} + 2 nπ$ , with $n \in Z$ .

The equation compares a sine with a cosine, so it is not yet of the form $sin A = sin B$ . With the related angle $sin (\frac{π}{2} - x) = cos x$ the right-hand side becomes a sine, and from there one proceeds with the usual two alternatives. In the second alternative the argument $π - (\frac{π}{2} - x)$ simplifies to $\frac{π}{2} + x$ , and the resulting equation is linear in $x$ .

The same result is obtained another way: with the double-angle formula $sin 2 x = 2 sin x cos x$ the equation becomes $cos x (2 sin x - 1) = 0$ , which gives $cos x = 0$ , that is $x = \frac{π}{2} + nπ$ , or $sin x = \frac{1}{2}$ , that is $x = \frac{π}{6} + 2 nπ$ or $x = \frac{5 π}{6} + 2 nπ$ . Over one turn both approaches give the angles $\frac{π}{6}$ , $\frac{π}{2}$ , $\frac{5 π}{6}$ and $\frac{3 π}{2}$ , therefore the two solution sets coincide.