# solve $\sin(3x - 4) = \cos(7x)$

I am attempting to solve the equation $$\sin(3x + 4) = \cos(7x),$$ with all numbers in degrees.

My process is as such:

$\cos(7x) = \sin(90 - 7x)$

$\sin(3x + 4) = \sin(90 - 7x)$

$3x + 4 = 90 - 7x + 360n$; (where n is an integer and $360n$ is added due to cycling)

$x = \frac{94}{10} + 36n$

However, when I graph, I see that there is another answer that I have not solved for, which is $\frac{133}{2} + 90n$. How can I achieve this answer?

• See, $90−4$ does not equal $94$... And equality of sines has one more solution: $\sin(a)=\sin(b)\implies (a=b+2n180^\circ \lor a=-b+(2n+1)180^\circ)$, because $\sin(x)=\sin(360^\circ+x)$ and $\sin(x)=\sin(180^\circ−x)$. Commented Nov 25, 2015 at 8:19

If $\sin(A)=\sin(B)$ then either

$A = B+360n$ or $A=180-B+360n$ where $n\in Z$

For example

$\sin(30)=\sin(330)=\frac{1}{2}$ and $\sin(30)=\sin(150)=\frac{1}{2}$

So basically you choose one of the two interior functions as the reference angle and use it in a similar way than normal general solutions

Draw a graph of $\sin x$.

You will quickly see that $\sin a = \sin b$ does not imply $a \equiv b \pmod{360}$ - just see $\sin 60 = \sin 120$.

The rest is left to the reader.

$$\sin(3x+4)=\sin3x\cos4+\cos3x\sin4$$ and \begin{align}\cos(7x)+\cos90&=2\cos\frac{(90+7x)}{2}.\cos\frac{(90-7x)}{2}\\ &=\cos^2(45)-\sin^2(7x/2)\\ &=\frac{1}{4}-\sin^2(7x/2)\end{align} now you have two sin equations can you the graphs from here if you think cos and sin are bit difficult.

• I do not see how this helps OP solve the problem. Commented Nov 25, 2015 at 12:54
• Sir i think its easy to compare graphs of sine's. Commented Nov 25, 2015 at 12:56