# (sin 3x + sin x) / (cos 3x + cos x) = sqrt 3, sin x = ?

Feb 25, 2018

$\sin x = \frac{1}{2}$ [when $0 \le x \le \frac{\pi}{2}$]

#### Explanation:

$\frac{\sin 3 x + \sin x}{\cos 3 x + \cos x} = \sqrt{3}$

We'll apply 2 identities here, $\textcolor{red}{\sin 3 x = 3 \sin x - 4 {\sin}^{3} x}$ and $\textcolor{red}{\cos 3 x = 4 {\cos}^{3} x - 3 \cos x}$

or, $\frac{\textcolor{red}{3 \sin x - 4 {\sin}^{3} x} + \sin x}{\textcolor{red}{4 {\cos}^{3} x - 3 \cos x} + \cos x} = \sqrt{3}$

Combining $\sin x$ terms,

or, $\frac{4 \sin x - 4 {\sin}^{3} x}{4 {\cos}^{3} x - 2 \cos x} = \sqrt{3}$

or, $\frac{4 \sin x \left(1 - {\sin}^{2} x\right)}{2 \cos x \left(2 {\cos}^{2} x - 1\right)} = \sqrt{3}$

Again, 2 more identities, color(magenta)(cos^2x = 1-sin^2x and color(magenta)(cos2x=2cos^2x-1

or, $\frac{2 \sin x {\cancel{\textcolor{m a \ge n t a}{{\cos}^{2} x}}}^{\cos x}}{\cancel{\cos} x \textcolor{m a \ge n t a}{\cos 2 x}} = \sqrt{3}$

or, $\frac{2 \sin x \cos x}{\cos 2 x} = \sqrt{3}$

One more here, color(blue)(sin2x=2sinxcosx

or, $\frac{\textcolor{b l u e}{\sin 2 x}}{\cos 2 x} = \sqrt{3}$

or, $\tan 2 x = \sqrt{3}$

or, $\tan 2 x = \tan \left(\frac{\pi}{3}\right)$

or,$2 x = \frac{\pi}{3}$

or, $x = \frac{\pi}{6}$

so, $\sin x = \sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$

Feb 25, 2018

$\sin x = \frac{1}{2}$

#### Explanation:

We may use,

$\rightarrow \sin A + \sin B = 2 \sin \left(\frac{A + B}{2}\right) \cdot \cos \left(\frac{A + B}{2}\right)$ and

$\rightarrow \cos A + \cos B = 2 \cos \left(\frac{A + B}{2}\right) \cdot \cos \left(\frac{A + B}{2}\right)$

Given that,

$\rightarrow \frac{\sin 3 x + \sin x}{\cos 3 x + \cos} = \sqrt{3}$

$\rightarrow \frac{\cancel{2} \sin \left(\frac{3 x + x}{2}\right) \cdot \cancel{\cos \left(\frac{3 x - x}{2}\right)}}{\cancel{2} \cos \left(\frac{3 x + x}{2}\right) \cdot \cancel{\cos \left(\frac{3 x - x}{2}\right)}} = \sqrt{3}$

rarrtan2x=tan60°

rarr2x=60°

rarrx=30°

Now, sinx=sin30°=1/2