# Thread: Hence what? rsin(x+alpha) and solving a trig equation.

1. ## Hence what? rsin(x+alpha) and solving a trig equation.

Hiya, either I'm missing something or my book is being weird:
Q) Express $4\sin x + 3\cos x$ in the form $r\sin(x+\alpha)$.
Hence find all the values of x in the range $0\le x \le 360^\circ$ for which $\cos 3x=\cos 2x$
Q END)

I can get the $r\sin(x+\alpha)$ form of $4\sin x + 3\cos x$:
$5\sin(x+36.9^\circ)$, but I don't see how that helps me with the second part of the question.

I think I can do the second part of the question anyway. By using $\cos 3x= \cos(2x + x)$ and then using substitutions to change all 2x into single x and to get rid of all the sinx that appear, I get:
$4\cos^3 x -2\cos^2 x -3\cos x + 1 = 0$
Then subsituting $y=\cos x$ and solving for y gives $(y-1)(4y^2 +2y-1) \Rightarrow y=1, y=\frac{-1\pm\sqrt{5}}{4}$
$\Rightarrow x=0^\circ,360^\circ,72^\circ,288^\circ,144^\circ,2 16^\circ$

But the book gives as an answer: $119.6^\circ$ and $346.7^\circ$
I made the graph and I think it backs up my answer, the brown line has roots where the blue and green lines cross. So why on earth do they say 'hence' as if the first part of the question should help solve the second part? Is the question just fubared or am I missing something(s)?
Thanks.

2. Hello tleave2000
Originally Posted by tleave2000
Hiya, either I'm missing something or my book is being weird:
Q) Express $4\sin x + 3\cos x$ in the form $r\sin(x+\alpha)$.
Hence find all the values of x in the range $0\le x \le 360^\circ$ for which $\cos 3x=\cos 2x$
Q END)

I can get the $r\sin(x+\alpha)$ form of $4\sin x + 3\cos x$:
$5\sin(x+36.9^\circ)$, but I don't see how that helps me with the second part of the question.
I must say, I don't see that either. But the easiest way to solve $\cos3x =\cos2x$ is as follows:

The general solution of $\cos A = \cos B$ is

$A = 2n\pi \pm B, n = 0, \pm1, \pm2, ...$

(Check out the graph of $y = \cos x$ if you want to confirm this.)

So $\cos3x =\cos2x \Rightarrow 3x = 2n\pi \pm 2x$

Taking the plus sign on the RHS gives $x = 2n\pi$, and taking the minus sign gives $5x = 2n\pi$; i.e. $x = \tfrac25n\pi$

So in the range $0\le x \le 2\pi$, the solutions are:

$x = 0, \tfrac25\pi, \tfrac45\pi, \tfrac65\pi, \tfrac85\pi, 2\pi$

Or, in degrees, the solutions you found: $0^o,72^o,144^o,216^o,288^o,360^o$

The book's answers are clearly wrong: they just don't give $\cos3x = \cos2x$. So, I'm on your side here! I suspect someone has got some wires crossed somewhere.