Hiya, either I'm missing something or my book is being weird:

Q) Express $\displaystyle 4\sin x + 3\cos x$ in the form $\displaystyle r\sin(x+\alpha)$.

Hence find all the values of x in the range $\displaystyle 0\le x \le 360^\circ$ for which $\displaystyle \cos 3x=\cos 2x$

Q END)

I can get the $\displaystyle r\sin(x+\alpha)$ form of $\displaystyle 4\sin x + 3\cos x$:

$\displaystyle 5\sin(x+36.9^\circ)$, 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 $\displaystyle \cos 3x= \cos(2x + x)$ and then using substitutions to change all 2x into single x and to get rid of all the sinx that appear, I get:

$\displaystyle 4\cos^3 x -2\cos^2 x -3\cos x + 1 = 0$

Then subsituting $\displaystyle y=\cos x$ and solving for y gives $\displaystyle (y-1)(4y^2 +2y-1) \Rightarrow y=1, y=\frac{-1\pm\sqrt{5}}{4}$

$\displaystyle \Rightarrow x=0^\circ,360^\circ,72^\circ,288^\circ,144^\circ,2 16^\circ$

But the book gives as an answer: $\displaystyle 119.6^\circ$ and $\displaystyle 346.7^\circ$

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.