Solve-for-x-in-R-2-sin-3x-4-3-0-

Question Number 49765 by hassentimol last updated on 10/Dec/18

Solve for x in R :

2 \times \sin (3 x + 4) + \sqrt{3} = 0

Commented by mathmax by abdo last updated on 27/Feb/20

(e) \Leftrightarrow s i n (3 x + 4) = - \frac{\sqrt{3}}{2} \Rightarrow s i n (3 x + 4) = s i n (- \frac{π}{3}) \Rightarrow

3 x + 4 = - \frac{π}{3} + 2 k π o r 3 x + 4 = \frac{4 π}{3} + 2 k π \Rightarrow 3 x = - 4 - \frac{π}{3} + 2 k π o r

3 x = - 4 + \frac{4 π}{3} + 2 k π \Rightarrow x = - \frac{4}{3} - \frac{π}{9} + \frac{2 k π}{3} o r x = - \frac{4}{3} - \frac{π}{9} + \frac{2 k π}{3}

Answered by afachri last updated on 10/Dec/18

2 \sin (3 x + 4) = - \sqrt{3^{}} \Rightarrow \sin (3 x + 4) = - \frac{\sqrt{3^{}}}{2}

consider 3 x + 4 = y, so : \sin (y) = - \sin (\frac{π}{3})

Sin has negative value in 3^{rd} and 4^{th} quadrant . p

∙ 3^{rd} quadrant \Rightarrow \sin (π + y) = - \sin (\frac{π}{3})

y = π + \frac{π}{3} = \frac{4 π}{3}

∙ 4^{th} quadrant \Rightarrow \sin (2 π - y) = - \sin (\frac{π}{3})

y = 2 π - \frac{π}{3} = \frac{5 π}{3}

\frac{4 π}{3} = 3 x + 4 \Rightarrow x_{1} = \frac{4 π - 12}{9}

\frac{5 π}{3} = 3 x + 4 \Rightarrow x_{2} = \frac{5 π - 12}{9}

Answered by mr W last updated on 10/Dec/18

\sin (3 x + 4) = - \frac{\sqrt{3}}{2}

\Rightarrow 3 x + 4 = 2 n π - \frac{π}{3} o r (2 n + 1) π + \frac{π}{3}

\Rightarrow x = \frac{(6 n - 1) π}{9} - \frac{4}{3} o r \frac{2 (3 n + 2) π}{9} - \frac{4}{3}

w i t h n = 0, \pm 1, \pm 2, \dots

Commented by afachri last updated on 10/Dec/18

did i solve wrong sir ? ? ? please let me know

Sir :) hope you {don}^{'} t mind

Commented by mr W last updated on 10/Dec/18

y o u r s o l u t i o n i s n o t c o m p l e t e s i r .

y o u t a k e o n l y 3 x + 4 = π + \frac{π}{3}, b u t i t c a n a l s o

b e 3 π + \frac{π}{3}, 5 π + \frac{π}{3}, e t c . - π + \frac{π}{3}, - 3 π + \frac{π}{3}, e t c .

y o u t a k e o n l y 3 x + 4 = 2 π - \frac{π}{3}, b u t i t c a n a l s o

b e 4 π - \frac{π}{3}, 6 π - \frac{π}{3}, e t c . - \frac{π}{3}, - 2 π + \frac{π}{3}, e t c .

a s t h e q u e s t i o n s a y s x \in R, i . e . - \infty < x < + \infty .

Commented by hassentimol last updated on 10/Dec/18

Sir, at the 2 nd line, {isn}^{'} t it 2 n π - \frac{2 π}{3} instead

for the second one \dots ?

Commented by afachri last updated on 10/Dec/18

thanks a lot for your correction and explanation

Mr . W, appreciate that . i^{'} ll fix my job again .

h ope you never be bored Sir . :) :)

Commented by mr W last updated on 10/Dec/18

2 n π - \frac{2 π}{3} i s t h e s a m e a s (2 n + 1) π + \frac{π}{3}

b e c a u s e n i s a n y i n t e g e r h e r e .

Commented by hassentimol last updated on 10/Dec/18

Ok \dots thanks sir I^{'} ve forgotten it \dots