Solution 4.4:7c

From Förberedande kurs i matematik 1

If we want to solve the equation cos 3x=sin 4x, we need an additional result which tells us for which values of u and v the equality cos u=sin v holds, but to get that we have to start with the equality cosu=cosv.

So, we start by looking at the equality

cosu=cosv

We know that for fixed u there are two angles v=u and v=−u in the unit circle which have the cosine value cosu, i.e. their x -coordinate is equal to cosu.

Imagine now that the whole unit circle is rotated anti-clockwise an angle 2. The line x=cosu will become the line y=cosu and the angles u and \displaystyle -u are rotated to \displaystyle u+{\pi }/{2}\; and \displaystyle -u+{\pi }/{2}\;, respectively.

The angles \displaystyle u+{\pi }/{2}\; and \displaystyle -u+{\pi }/{2}\; therefore have their \displaystyle y -coordinate, and hence sine value, equal to \displaystyle \cos u. In other words, the equality

\displaystyle \text{cos }u=\text{sin }v

holds for fixed \displaystyle u in the unit circle when \displaystyle v=\pm u+{\pi }/{2}\;, and more generally when

\displaystyle v=\pm u+\frac{\pi }{2}+2n\pi ( \displaystyle n an arbitrary integer).

For our equation \displaystyle \text{cos 3}x=\text{sin 4}x, this result means that \displaystyle x\text{ } must satisfy

\displaystyle 4x=\pm 3x+\frac{\pi }{2}+2n\pi

This means that the solutions to the equation are

\displaystyle \left\{ \begin{array}{*{35}l} x=\frac{\pi }{2}+2n\pi \\ x=\frac{\pi }{14}+\frac{2}{7}\pi n \\ \end{array} \right. ( \displaystyle n an arbitrary integer)