# Solve (2+sqrt3)cos theta=1-sin theta?

## $\left(2 + \sqrt{3}\right) \cos \theta = 1 - \sin \theta$

Mar 10, 2018

$\rightarrow x = \left(6 n - 1\right) \cdot \left(\frac{\pi}{3}\right)$

$\rightarrow x = \left(4 n + 1\right) \frac{\pi}{2}$ Where $n \rightarrow Z$

#### Explanation:

$\rightarrow \left(2 + \sqrt{3}\right) \cos x = 1 - \sin x$

$\rightarrow \tan {75}^{\circ} \cdot \cos x + \sin x = 1$

$\rightarrow \frac{\sin {75}^{\circ} \cdot \cos x}{\cos {75}^{\circ}} + \sin x = 1$

$\rightarrow \sin x \cdot \cos {75}^{\circ} + \cos x \cdot \sin {75}^{\circ} = \cos {75}^{\circ} = \sin \left({90}^{\circ} - {15}^{\circ}\right) = \sin {15}^{\circ}$

$\rightarrow \sin \left(x + {75}^{\circ}\right) - \sin {15}^{\circ} = 0$

$\rightarrow 2 \sin \left(\frac{x + {75}^{\circ} - {15}^{\circ}}{2}\right) \cos \left(\frac{x + {75}^{\circ} + {15}^{\circ}}{2}\right) = 0$

$\rightarrow \sin \left(\frac{x + {60}^{\circ}}{2}\right) \cdot \cos \left(\frac{x + {90}^{\circ}}{2}\right) = 0$

Either $\rightarrow \sin \left(\frac{x + {60}^{\circ}}{2}\right) = 0$

$\rightarrow \frac{x + {60}^{\circ}}{2} = n \pi$

$\rightarrow x = 2 n \pi - {60}^{\circ} = 2 n \pi - \frac{\pi}{3} = \left(6 n - 1\right) \cdot \left(\frac{\pi}{3}\right)$

or, $\cos \left(\frac{x + {90}^{\circ}}{2}\right) = 0$

$\rightarrow \frac{x + {90}^{\circ}}{2} = \left(2 n + 1\right) \frac{\pi}{2}$

$\rightarrow x = 2 \cdot \left(2 n + 1\right) \frac{\pi}{2} - \frac{\pi}{2} = \left(4 n + 1\right) \frac{\pi}{2}$

Mar 10, 2018

If, $\cos \theta = 0 \implies \sin \theta = 1 \implies \theta = \left(4 k + 1\right) \frac{\pi}{2} , k \in Z$
$\theta = 2 k \pi - \frac{\pi}{3} , k \in Z$,

#### Explanation:

$\left(2 + \sqrt{3}\right) \cos \theta = 1 - \sin \theta$
$\mathmr{and} \cos \theta \ne 0$, dividing both sides by $\cos \theta$
$2 + \sqrt{3} = \sec \theta - \tan \theta \implies \sec \theta - \tan \theta = 2 + \sqrt{3} \to \left(I\right)$
$\therefore \frac{1}{\sec \theta - \tan \theta} = \frac{1}{2 + \sqrt{3}}$$\implies \frac{{\sec}^{2} \theta - {\tan}^{2} \theta}{\sec \theta - \tan \theta} = \frac{1}{2 + \sqrt{3}} \cdot \frac{2 - \sqrt{3}}{2 - \sqrt{3}}$
$\implies \sec \theta + \tan \theta = 2 - \sqrt{3} \to \left(I I\right)$
Adding $\left(I\right) \mathmr{and} \left(I I\right)$,we get.$2 \sec \theta = 4 \implies \sec \theta = 2$
$\textcolor{red}{\cos \theta = \frac{1}{2} > 0}$, From the given equn.
$\cos \theta = \frac{1}{2} \implies \left(2 + \sqrt{3}\right) \left(\frac{1}{2}\right) = 1 - \sin \theta$$\implies 1 + \frac{\sqrt{3}}{2} = 1 - \sin \theta \implies \textcolor{red}{\sin \theta = - \frac{\sqrt{3}}{2} < 0}$
$\theta = 2 k \pi - \frac{\pi}{3} , k \in Z$,.......... (IV^(th)quadrant)