# Solve the equation 5sin4x=2sin2x if 0^@ <= x <= 360^@?

Oct 17, 2016

Solution is $\left\{{0}^{o} , {39.232}^{o} , {90}^{o} , {140.768}^{o} , {180}^{o} , {219.232}^{o} , {270}^{o} , {320.768}^{o} , {360}^{o}\right\}$

#### Explanation:

$5 \sin 4 x = 2 \sin 2 x$

$\Leftrightarrow 5 \times 2 \sin 2 x \cos 2 x = 2 \sin 2 x$

or $10 \sin 2 x \cos 2 x - 2 \sin 2 x = 0$

or $2 \sin 2 x \left(5 \cos 2 x - 1\right) = 0$

Hence either $\sin 2 x = 0$ i.e. $2 x = \left\{{0}^{o} , {180}^{o} , {360}^{o} , {540}^{o} , {720}^{o} , \ldots .\right\}$

or in the given range $0 \le x \le {360}^{o}$, $x = \left\{{0}^{o} , {90}^{o} , {180}^{o} , {270}^{o} , {360}^{o}\right\}$

or $5 \cos 2 x - 1 = 0$ i.e. $\cos 2 x = \frac{1}{5} = \cos \left(\pm {78.463}^{o}\right)$ i.e.

$2 x = \left\{{78.463}^{o} , \left({360}^{o} - {78.463}^{o}\right) , \left({360}^{o} + {78.463}^{o}\right) , \left({720}^{o} - {78.463}^{o}\right)\right\}$

$x = \left\{{39.232}^{o} , {140.768}^{o} , {219.232}^{o} , {320.768}^{o}\right\}$

Hence, solution is $\left\{{0}^{o} , {39.232}^{o} , {90}^{o} , {140.768}^{o} , {180}^{o} , {219.232}^{o} , {270}^{o} , {320.768}^{o} , {360}^{o}\right\}$
graph{5sin(4x)-2sin(2x) [-1, 7.26, -4, 6]}