# Solving $\cos(x)\sin(7x)=\cos(3x)\sin(5x)$

Recently, I was trying to solve a trigonometric equation involving the use of sine and cosine:

$$\cos(x)\sin(7x)=\cos(3x)\sin(5x)$$

I attempted to remove the coefficients of $$x$$ within the trigonometric functions, but I did not understand how to.

Does anyone understand how I'm supposed to solve this equation?

Thank you!

Hint:

Use the linearisation formula: $$\sin a\cos b=\frac12\bigl(\sin(a+b)+\sin(a-b)\bigr).$$

Hint: there is a set of identities called by names such as product rules or product-to-sum formulas. One of them has $$\sin(a)\cos(b)$$ on the left-hand side.

Multiply both sides by $$2,$$ and use the identity $$2\cos x\sin y=\sin (x+y)-\sin(x-y).$$ This gives $$\sin 8x-\sin(-6x)=\sin 8x-\sin(-2x),$$ or in other terms $$\sin 6x-\sin 2x=0.$$ Here you may convert this into a product by using $$\sin a-\sin b=2\cos\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right),$$ to get a product that vanishes, then use the fact that if $$AB=0,$$ then $$A=0$$ or $$B=0$$ to finish off the problem.

PS. Note that if $$\cos m=0,$$ then $$m=(1+2j)π/2,$$ and if $$\sin n=0,$$ then $$n=kπ,$$ where $$j,k$$ are arbitrary integers.

An alternate path is to use,

$$\cos(x) = \frac{1}{2}(e^{ix} + e^{-ix})$$ and $$\sin(x) = \frac{1}{2i}(e^{ix} - e^{-ix})$$. Let $$z=e^{ix}$$ and keep in mind that $$|z|=1$$. After some algebra you will get the polynomial,

$$z^{14} - z^{10} + z^6 - z^2 = 0$$

Which factors as

$$z^2(z^8 + 1)(z^4 - 1) = 0$$

Since $$|z| = 1$$ we can ignore the $$z^2$$. Then we have solutions when either $$e^{i8x} = -1$$ or $$e^{i4x} = 1$$. The first equation is satisfied when $$8x = (1 + 2k)\pi$$ (an odd integer times pi) and the second is satisfied when $$4x = 2k\pi$$ (an even integer times pi).