# How do you solve cosx(3sinx - 5) = 0?

Feb 1, 2016

#### Answer:

$\left(2 n + 1\right) \frac{\pi}{2}$[where $n \in \mathbb{Z}$]

#### Explanation:

here,
$\cos x \left(3 \sin x - 5\right) = 0$

$\mathmr{and} , 3 \sin x \cos x - 5 \cos x = 0$

$\mathmr{and} , 3 \cdot 2 \sin x \cos x = 5 \cdot 2 \cos x$

$\mathmr{and} , 3 \sin 2 x = 10 \cos x$

$\mathmr{and} , 9 {\sin}^{2} 2 x = 100 {\cos}^{2} x$[taking square on each sides]

$\mathmr{and} , 9 \cdot 2 {\sin}^{2} 2 x = 100 \cdot 2 {\cos}^{2} x$

$\mathmr{and} , 18 \left(1 - {\cos}^{2} 2 x\right) = 100 \left(1 + \cos 2 x\right)$

$\mathmr{and} , 18 - 18 {\cos}^{2} 2 x = 100 + 100 \cos 2 x$

$\mathmr{and} , 18 {\cos}^{2} 2 x + 100 \cos 2 x + 82 = 0$

$\mathmr{and} , 18 {\cos}^{2} 2 x + 18 \cos 2 x + 82 \cos 2 x + 82 = 0$

$\mathmr{and} , 18 \cos 2 x \left(\cos 2 x + 1\right) + 82 \left(\cos 2 x + 1\right) = 0$

$\mathmr{and} , \left(\cos 2 x + 1\right) \left(18 \cos 2 x + 82\right) = 0$

one solution is,

$\cos 2 x + 1 = 0$

$\mathmr{and} , \cos 2 x = - 1$

$\mathmr{and} , 2 x = \left(2 n + 1\right) \pi$[where, $n \in \mathbb{Z}$]

$\mathmr{and} , x = \left(2 n + 1\right) \frac{\pi}{2}$

another solution is,

$18 \cos 2 x + 82 = 0$

$\mathmr{and} , \cos 2 x = - \frac{82}{18}$

but, $\cos 2 x \ne - \frac{82}{18}$ [as, $\cos 2 x \cancel{>} 1$]

so, $x = \left(2 n + 1\right) \frac{\pi}{2}$[where $n \in \mathbb{Z}$]