# Number of solutions of sinx.tan4x=cosx in (0,π) ?

Jun 23, 2018

$x \in \left\{\frac{\pi}{10} , 3 \frac{\pi}{10} , \frac{\pi}{2} , 7 \frac{\pi}{10} , 9 \frac{\pi}{10}\right\} \subset \left(0 , \pi\right)$.

#### Explanation:

Note that, in the given eqn., $\sin x \ne 0 ,$ because, otherwise

$\cos x = \sin x \tan 4 x = 0 , \text{ contradicting, } {\sin}^{2} + {\cos}^{2} x = 1.$

Hence, we can divide the eqn. by $\sin x ,$ and get,

$\tan 4 x = \cos \frac{x}{\sin} x = \cot x = \tan \left(\frac{\pi}{2} - x\right)$.

$\therefore 4 x = \left(\frac{\pi}{2} - x\right) + k \pi , i . e . , 5 x = \frac{\pi}{2} + k \pi , k \in \mathbb{Z}$.

$\therefore x = \frac{\pi}{10} + k \frac{\pi}{5} = \left(2 k + 1\right) \frac{\pi}{10} , k \in \mathbb{Z}$.

But, $0 < x < \pi \therefore 0 < \left(2 k + 1\right) \frac{\pi}{10} < \pi$

$\therefore 0 < 2 k + 1 < 10 , \mathmr{and} , - \frac{1}{2} < k < \frac{9}{2} , k \in \mathbb{Z}$.

$\therefore x = \left(2 k + 1\right) \frac{\pi}{10} , k = 0 , 1 , 2 , 3 , 4$.

$\therefore x \in \left\{\frac{\pi}{10} , 3 \frac{\pi}{10} , \frac{\pi}{2} , 7 \frac{\pi}{10} , 9 \frac{\pi}{10}\right\} \subset \left(0 , \pi\right)$.

Jun 23, 2018

Number of solutions: 5 for $\left(0 , \pi\right)$

#### Explanation:

sin x.tan 4x = cos x
$\tan 4 x = \cot x = \tan \left(\frac{\pi}{2} - x\right)$
Unit circle and property of tan give -->
$4 x = \left(\frac{\pi}{2} - x\right) + k \pi$
$5 x = \frac{\pi}{2} + k \pi$
$x = \frac{\pi}{10} + \frac{k \pi}{5}$
k = 0 --> $x = \frac{\pi}{10}$
k = 1 --> $x = \frac{\pi}{10} + \frac{\pi}{5} = \frac{3 \pi}{10}$
k = 2 --> $x = \frac{\pi}{10} + \frac{2 \pi}{5} = \frac{5 \pi}{10}$
k = 3 --> $x = \frac{\pi}{10} + \frac{3 \pi}{5} = \frac{7 \pi}{5}$
k = 4 --> $x = \frac{\pi}{10} + \frac{4 \pi}{5} = \frac{9 \pi}{10}$
In the interval $\left(0 , \pi\right)$, there are 5 answers.