# Question #abd69

Jun 21, 2016

$\left({72}^{\circ} 26 + k 180\right)$
$\left({220}^{\circ} + k 180\right)$

#### Explanation:

Apply these identities, calling $\tan \left(\frac{x}{2}\right) = t$
$\sin x = \frac{2 t}{1 + {t}^{2}}$
$\cos x = \frac{1 - {t}^{2}}{1 + {t}^{2}}$
2sin x - 3cos x = 1
Replace sin x and cos x by the above identities, we get:
$4 t - 3 + 3 {t}^{2} = 1 + {t}^{2}$
$2 {t}^{2} + 4 t - 4 = 0$
${t}^{2} + 2 t - 2 = 0$.
Solve this quadratic equation for t, using the improved quadratic formula (Google, Yahoo Search);
$D = {d}^{2} = {b}^{2} - 4 a c = 4 + 8 = 12$ --> $d = \pm 2 \sqrt{3}$
There are 2 real roots:
$t = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{2}{2} \pm \frac{2 \sqrt{3}}{2} = - 1 \pm \sqrt{3}$
$t 1 = - 1 - \sqrt{3} = - 2.73$
$t 2 = - 1 + \sqrt{3} = 0.73$
a. $t 1 = \tan \left(\frac{x}{2}\right) = - 2.73$ --> $\frac{x}{2} = - {69}^{\circ} 88$ and $\frac{x}{2} = - 69.88 + 180 = {110}^{\circ} 12$ --> x = 220.24 + k180
b. t2 = 0.73 --> $\frac{x}{2} = 36.13$ and $\frac{x}{2} = 36.13 + 180 = 216.13$ -->
$x = 72.26 + k 180$

Jun 21, 2016

Using identities of fractional angle

$\sin x = \frac{2 \tan \left(\frac{x}{2}\right)}{1 + {\tan}^{2} \left(\frac{x}{2}\right)}$

and
$\cos x = \frac{1 - {\tan}^{2} \left(\frac{x}{2}\right)}{1 + {\tan}^{2} \left(\frac{x}{2}\right)}$

Given equation becomes

$2 \sin x - 3 \cos x = 1$

$= \frac{4 \tan \left(\frac{x}{2}\right)}{1 + {\tan}^{2} \left(\frac{x}{2}\right)} - 3 \frac{\left(1 - {\tan}^{2} \left(\frac{x}{2}\right)\right)}{1 + {\tan}^{2} \left(\frac{x}{2}\right)} = 1$

putting $\tan \left(\frac{x}{2}\right) = t$ we get

$\implies \frac{4 t}{1 + {t}^{2}} - \frac{3 \left(1 - {t}^{2}\right)}{1 + {t}^{2}} = 1$

$\implies 4 t - 3 + 3 {t}^{2} = 1 + {t}^{2}$

$\implies 3 {t}^{2} - {t}^{2} + 4 t - 3 - 1 = 0$

$\implies 2 {t}^{2} + 4 t - 4 = 0$

$\implies {t}^{2} + 2 t - 2 = 0$

$\implies {t}^{2} + 2 t + 1 = 3$

$\implies {\left(t + 1\right)}^{2} = 3$

$\implies t + 1 = \pm \sqrt{3}$

$\implies t = \pm \sqrt{3} - 1$

when $t = \sqrt{3} - 1$

$\tan \left(\frac{x}{2}\right) = \sqrt{3} - 1$

$\therefore x = 2 \times {\tan}^{-} 1 \left(\sqrt{3} - 1\right) = 2 \cdot {36.2}^{\circ} = {72.4}^{\circ}$

Or,$x = \left(180 + 72.4\right) = {252.4}^{\circ}$

Again when

$\tan \left(\frac{x}{2}\right) = - \sqrt{3} - 1$

$\implies x = 2 \times {\tan}^{-} 1 \left(- \sqrt{3} - 1\right)$

$\implies x = 2 \cdot {110.1}^{\circ} = {220.2}^{\circ}$