1. ## Trigonometric equation

Solve $\sqrt{3}\cos{2x}-\sin{x}\cos{x}-1 = 0$ for $-2\pi \leq x \leq 2\pi$.

2. Hello Elvis
Originally Posted by Elvis
Solve $\sqrt{3}\cos{2x}-\sin{x}\cos{x}-1 = 0$ for $-2\pi \leq x \leq 2\pi$.
Using $\sin 2x = 2 \sin x \cos x$, write the equation as:
$\sqrt3\cos 2x -\tfrac12\sin2x = 1$
Then express the LHS in the form $r\cos(2x+\alpha)$, where:
$\left\{\begin{array}{l}
r\cos\alpha = \sqrt3\\
r\sin\alpha = \tfrac12\\
\end{array}\right.$

Can you continue?

Hello ElvisUsing $\sin 2x = 2 \sin x \cos x$, write the equation as:
$\sqrt3\cos 2x -\tfrac12\sin2x = 1$
Then express the LHS in the form $r\cos(2x+\alpha)$, where:
$\left\{\begin{array}{l}
r\cos\alpha = \sqrt3\\
r\sin\alpha = \tfrac12\\
\end{array}\right.$

Can you continue?

$\left\{\begin{array}{l}
r\cos\alpha = \sqrt3\\
r\sin\alpha = \tfrac12\\
\end{array}\right.$

We have $r\cos\alpha = \sqrt3 \Rightarrow \cos\alpha = \dfrac{\sqrt{3}}{r}$ And $r\sin\alpha = \dfrac{1}{2} \Rightarrow \sin\alpha = \dfrac{0.5}{r}$. Then $r = \sqrt{\left(\sqrt{3}\right)^2+\left(\dfrac{1}{2}\r ight)^2} = \dfrac{\sqrt{13}}{2}$ and $\alpha = \arctan{\left(\dfrac{0.5}{\sqrt{3}}\right)} = 16.10^{\circ}$. Hence $\sqrt3\cos 2x -\tfrac12\sin2x = \dfrac{\sqrt{13}}{2}\cos\left(2x+16.10^{\circ}\rig ht)$
But $\sqrt3\cos 2x -\tfrac12\sin2x = 1$ So $\dfrac{\sqrt{13}}{2}\cos\left(2x+16.10^{\circ}\rig ht) = 1 \Leftrightarrow \cos\left(2x+16.10^{\circ}\right) = \dfrac{2}{\sqrt{13}}$ $\Rightarrow 2x+16.10^{\circ} = \arccos{\left(\dfrac{2}{\sqrt{13}}\right)} = 56.30^{\circ}$ or $326.3^{\circ}$

$\Rightarrow 2x = 56.30^{\circ}-16.10^{\circ} = 40.2^{\circ}$, or $326.3^{\circ}-16.10^{\circ} = 310.2^{\circ}$ $\Rightarrow x = \frac{40.2^{\circ}}{2}$, or $\dfrac{310.2^{\circ}}{2}$ $\Rightarrow x = 20.1^{\circ}$ or $155.1^{\circ}$.

Hence the solutions of the equation $\sqrt{3}\cos{2x}-\sin{x}\cos{x}-1 = 0
$
in the interval $[-2\pi, 2\pi]$ are $x = \pm{20.1^{\circ}}$ and $x = \pm{155.1^{\circ}}$. Hope it's right.

4. Hello Elvis

Your working is correct down to here:
Originally Posted by Elvis

$\left\{\begin{array}{l}
r\cos\alpha = \sqrt3\\
r\sin\alpha = \tfrac12\\
\end{array}\right.$

We have $r\cos\alpha = \sqrt3 \Rightarrow \cos\alpha = \dfrac{\sqrt{3}}{r}$ And $r\sin\alpha = \dfrac{1}{2} \Rightarrow \sin\alpha = \dfrac{0.5}{r}$. Then $r = \sqrt{\left(\sqrt{3}\right)^2+\left(\dfrac{1}{2}\r ight)^2} = \dfrac{\sqrt{13}}{2}$ and $\alpha = \arctan{\left(\dfrac{0.5}{\sqrt{3}}\right)} = 16.10^{\circ}$. Hence $\sqrt3\cos 2x -\tfrac12\sin2x = \dfrac{\sqrt{13}}{2}\cos\left(2x+16.10^{\circ}\rig ht)$
But $\sqrt3\cos 2x -\tfrac12\sin2x = 1$ So $\dfrac{\sqrt{13}}{2}\cos\left(2x+16.10^{\circ}\rig ht) = 1 \Leftrightarrow \cos\left(2x+16.10^{\circ}\right) = \dfrac{2}{\sqrt{13}}$ $\Rightarrow 2x+16.10^{\circ} = \arccos{\left(\dfrac{2}{\sqrt{13}}\right)}$
But then you need to say:

$2x+16.10=360n\pm 56.30$

$\Rightarrow x +8.05 = 180n\pm28.15$

$\Rightarrow x = 180n\pm28.15-8.05$

Then taking $n = 0$ with the $+$ sign:
$x = 20.1^o$
$n = 0$ with the $-$ sign:
$x =-36.2^o$
... etc.

$x=-339.3^o, -216.2^o, -159.9^o, -36.2^o, 20.1^o, 143.8^o, 200.1^o,323.8^o$