Solution 4.4:7c

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

So, we start by looking at the equality


\displaystyle \cos u=\cos v


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


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