# Find, in radians the general solution of cos 3x = sin 5x

I am studying maths as a hobby. I have come across this problem:

Find a general solution for the equation cos 3x = sin 5x

I have said, $$\sin 5x = \cos(\frac{\pi}{2} - 5x)$$

so

$$\cos 3x = \sin 5x \implies 3x = 2n\pi\pm(\frac{\pi}{2} - 5x)$$

When I add $$(\frac{\pi}{2} - 5x)$$ to $$2n\pi$$ I get the answer $$x = \frac{\pi}{16}(4n +1)$$, which the book says is correct.

But when I subtract I get a different answer to the book. My working is as follows:

$$3x = 2n\pi - \frac{\pi}{2} + 5x$$

$$2x = \frac{\pi}{2} - 2n\pi$$

$$x = \frac{\pi}{4} - n\pi = \frac{\pi}{4}(1 - 4n)$$

but my text book says the answer is $$\frac{\pi}{4}(4n + 1)$$

Is the book wrong?

$$\sin 5x = \cos (\frac{\pi}{2}-5x)= \cos 3x$$

$$3x=\frac{\pi}{2}-5x+2k\pi$$

$$x=\frac{\pi}{16}+\frac{k\pi}{4}=\frac{\pi}{16}(1+4k)$$

or

$$3x=-(\frac{\pi}{2}-5x)+2k\pi$$

$$x=\frac{\pi}{4}-k\pi$$

$$x=\frac{\pi}{4}+k\pi =\frac{\pi}{4}(1+4k)$$

where $$k\in Z$$

writing $$-k\pi$$ or $$k\pi$$ does not change the solution set. Because $$-k$$ is the opposite of $$k$$ in integers.

• I'm not sure about the last sentence. I would have thought that because -k is the opposite of k then it must change the solution – Steblo Dec 3 '20 at 18:12
• I mean it will give the same solution set by writing $-k$ for $k$ – Lion Heart Dec 3 '20 at 18:15
• Is that because cos -x is the same as cosx? – Steblo Dec 3 '20 at 18:25
• @Steblo : take $x= \frac{\pi}{4}(1-4k)$ for $k=1$, $x=-\frac {3pi}{4}$. Now take take $x= \frac{\pi}{4}(1+4k)$ for $k=-1$, $x=-\frac {3pi}{4}$. – Lion Heart Dec 3 '20 at 18:32
• @Steblo : $\cos(-x)= \cos x$. Because Cosine function is an even function and its graph is symmetric to the y-axis or it can be say different reasons. – Lion Heart Dec 3 '20 at 18:43

If you write $$m$$ in place of $$n,$$ you reached at $$\dfrac{\pi(1-4m)}4$$

We $$\dfrac{\pi(1-4m)}4=\dfrac{\pi(1+4n)}4\iff m=-n$$

In our case $$m$$ is any integer $$\iff n=-m$$ also belong to the same infinite set of integers

In their case $$n$$ is so.

No, the two are equivalent. In particular, if $$m$$ = $$-n$$, then $$\dfrac{\pi}{2}(1 - 4m) = \dfrac{\pi}{2}(4n + 1),$$ so all that's really happened is tha tyou've listed the solutions in a different order.