# Thread: Solve for x: cos x + cos 3x = sin x + sin 3x

1. ## Solve for x: cos x + cos 3x = sin x + sin 3x

I copied down the solutions from my teacher for this question and i understand up to this point:

⇒ 0 = sin 6x - sin 2x

The next step it is:

⇒ 0 = 2cos ((6x+2x)/2) sin ((6x-2x)/2)

I don't understand how that the addition formulas were used to get to this step.

The solution i copied down is below:

cos x + cos 3x = sin x + sin 3x
⇒cos x - sin x = sin 3x - cos 3x
now, squaring on both sides,
(cos x - sin x)^2 = (sin 3x - cos 3x)^2
⇒(cos x)^2 – 2 sin x cos x + (sin x)^2 = (sin 3x)^2 - 2sin 3x cos 3x+ (cos 3x)^2
⇒ 1-2 sin x cos x = 1-2sin 3x cos 3x
⇒ -2 sin x cos x = -2sin 3x cos 3x
⇒ sin x cos x = sin 3x cos 3x
multiplying both sides by 2,
⇒ 2sin x cos x = 2sin 3x cos 3x
⇒ sin 2x = sin 2(3x)
⇒ sin 2x = sin 6x
⇒ 0 = sin 6x - sin 2x
⇒ 0 = 2cos ((6x+2x)/2) sin ((6x-2x)/2)
⇒ 0 = 2cos 4x sin 2x
⇒ 0 = cos 4x sin 2x
case 1:
⇒ cos 4x = 0
⇒ cos 4x = cos π/2
⇒ 4x = 2nπ + π/2
⇒ x = nπ/2 + π/8
case 2:
⇒ sin 2x = 0
⇒ sin 2x = sin π
⇒ 2x = nπ + ((-1)^n)(π)
⇒ x = nπ/2 + ((-1)^n)(π)/2

⇒ x = [nπ/2 + π/8] or [nπ/2 + ((-1)^n)(π)/2]

2. see this graph, only 6 solutions. figure out why?

3. Hello, BuffaloSoulja!

There are four sum-to-product identities:

. . $\begin{array}{cccc}(1) & \sin A + \sin B &=& 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) \\ \\[-3mm]
(2) & \sin A - \sin B &=& 2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) \\ \\[-3mm]
(3) & \cos A + \cos B &=&2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{ A-B}{2}\right) \\ \\[-3mm]
(4) & \cos A - \cos B &=& \text{-}2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)

\end{array}$

Your teacher's solution is rather silly . . .

He/she used (2) in that mysterious step.

What makes it silly is that he/she could have used these identities first
. . and save a lot of work.

$\cos x + \cos 3x \:=\: \sin x + \sin 3x$
We have: . $\cos3x + \cos x \:=\:\sin3x + \sin x$

Using (1) and (3), we have: . $2\cos2x\cos x \;=\;2\sin2x\cos x \quad\Rightarrow\quad 2\cos2x\cos x - 2\sin2x\sin x \;=\;0$

Factor: . $2\cos x(\cos2x - \sin 2x) \;=\;0$

And we have two equations to solve:

$2\cos x \:=\:0 \quad\Rightarrow\quad \cos x \:=\:0 \quad\Rightarrow\quad\boxed{ x \:=\:\frac{\pi}{2} + \pi n}$

$\cos2x\;-\;\sin2x\:=\:0 \quad\Rightarrow\quad \sin2x\:=\:\cos2x \quad\Rightarrow\quad \frac{\sin2x}{\cos2x} \:=\:1 \quad\Rightarrow\quad \tan2x \:=\:1$

. . $2x \:=\:\frac{\pi}{4} + \pi n \quad\Rightarrow\quad\boxed{ x \:=\:\frac{\pi}{8} + \frac{\pi}{2}n}$

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