# Math Help - Trig equation check

1. ## Trig equation check

I would like to know if my answer is correct

Q: Solve this trig equation where x is an element of 0, 360.

$\frac{cos2x-cos5x}{sin2x+sin5x}=2\sqrt{3}$

$\frac{-2sin\frac{1}{2}(7x)sin\frac{1}{2}(-3x)}{2sin\frac{1}{2}(7x)cos\frac{1}{2}(-3x)}=2\sqrt{3}$

$\frac{-2sin\frac{1}{2}(7x)}{2sin\frac{1}{2}(7x)}$ cancels out

leaving $-\frac{sin\frac{1}{2}(-3x)}{cos\frac{1}{2}(-3x)}=2\sqrt{3}$

which is $tan\frac{3x}{2}=-2\sqrt{3}$

$\frac{3x}{2}=106, 286$ (286 is out of range, of $\frac{3x}{2}$ so it gets removed)

$3x=112$

$\therefore x=37$

Thanks

2. ## I might be wrong but..

I don´t really understand the first step you have done?
$112/3 = 27,\frac{1}{3}$ (but I guess you know that).

Do you get $2\sqrt{3}$ when u substitute 37,3 back?
I got something different (I might be wrong)..

3. Hello, RAz!

Solve, where $x \in (0^o,360^o)\!:\;\;\frac{\cos2x-\cos5x}{\sin2x+\sin5x}\:=\:2\sqrt{3}$

We have: . $\frac{-2\sin\left(\dfrac{7x}{2}\right)\sin\left(-\dfrac{3x}{2}\right) }{2\sin\left(\dfrac{7x}{2}\right)\cos\left(-\dfrac{3x}{2}\right)} \;=\;2\sqrt{3} \qquad\Rightarrow\qquad \frac{\sin\left(\dfrac{7x}{2}\right)\sin\left(\dfr ac{3x}{2}\right)}{\sin\left(\dfrac{7x}{2}\right)\c os\left(\dfrac{3x}{2}\right)} \;=\;2\sqrt{3}$

. . . . . . . . . $\frac{\sin\left(\dfrac{3x}{2}\right)}{\cos\left(\d frac{3x}{2}\right)} \;=\;2\sqrt{3} \qquad\Rightarrow\qquad \tan\left(\frac{3x}{2}\right) \;=\;2\sqrt{3}$

Then: . $\frac{3x}{2} \;\approx\;73.89^o,\;253.89^o,\;433.89^o$

Therefore: . $x \;\approx\;49.26^o,\;169.26^o,\;289.26^o$