solve-in-pi-pi-E-sin3x-sin2x-

Question Number 81057 by mathocean1 last updated on 09/Feb/20

s o l v e in [- π; π]

(E) : s i n 3 x = - s i n 2 x

Commented by jagoll last updated on 09/Feb/20

\sin 3 x + \sin 2 x = 0

2 \sin (\frac{5 x}{2}) \cos (\frac{x}{2}) = 0

(1) \sin (\frac{5 x}{2}) = 0 = \sin 0

\frac{5 x}{2} = 2 n π \Rightarrow x = \frac{4 n π}{5}

(2) \cos (\frac{x}{2}) = 0 \Rightarrow \frac{x}{2} = \pm \frac{π}{2} + 2 n π

x = \pm π + 4 n π

Commented by mr W last updated on 09/Feb/20

i f \sin α = - \sin β, i t m e a n s

α = (2 k + 1) π + β o r α = 2 k π - β

s i n 3 x = - s i n 2 x

3 x = (2 k + 1) π + 2 x o r 3 x = 2 k π - 2 x

x = (2 k + 1) π o r x = \frac{2}{5} k π

w i t h i n [- π, π] w e h a v e

x = - π, π, - \frac{4 π}{5}, \frac{2 π}{5}, 0, \frac{2 π}{5}, \frac{4 π}{5}

Commented by jagoll last updated on 09/Feb/20

i f \sin α = \cos β ?

Commented by mr W last updated on 09/Feb/20

α + β = 2 k π + \frac{π}{2} o r β - α = 2 k π - \frac{π}{2}

i . e . β = 2 k π \pm (\frac{π}{2} - α)

y o u c a n s e e a l l t h e s e o n t h e g r a p h o f

t h e f u n c t i o n s .

Commented by jagoll last updated on 09/Feb/20

t h e s a m e c a s e f o r \sin α = - \cos β s i r

Commented by mr W last updated on 09/Feb/20

y o u c a n f i g u r e o u t a l l c a s e s b y y o u r s e l f

a c c o r d i n g t o t h e g r a p h o f t h e f u n c t i o n s .

\sin α = - \cos β \Rightarrow \sin (- α) = \cos β

u s i n g t h a t w h a t w e a l r e a d y k n o w a b o v e,

w e g e t

\Rightarrow β = 2 k π \pm (\frac{π}{2} + α)

Commented by john santu last updated on 09/Feb/20

\sin 3 x = \sin (- 2 x)

3 x = - 2 x + 2 k π

5 x = 2 k π \Rightarrow x = \frac{2 k π}{5}

o r 3 x = π + 2 x + 2 k π

x = π + 2 k π