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sin 2x = sin x and 0 ≤ x ≤ 2π. .....51°
note : sina = sinb equivaut : (a = b +2kπ) or ( a = π - b +2kπ) .... k∈ Z
(1) equiv : ( 2x = x +2kπ) or (2x = π - x +2kπ)
( x = 2kπ) or ( x = π/3 + 2kπ/3)....(general solution )
but : 0 ≤ x ≤ 2π
k = 0 : x = 0 , x = π/3
k= 1 : x = 2π , x = π
k = 2 ; x = 5π/3
sin 2x = sin x and 0 ≤ x ≤ 2π. .....51°
note : sina = sinb equivaut : (a = b +2kπ) or ( a = π - b +2kπ) .... k∈ Z
(1) equiv : ( 2x = x +2kπ) or (2x = π - x +2kπ)
( x = 2kπ) or ( x = π/3 + 2kπ/3)....(general solution )
but : 0 ≤ x ≤ 2π
k = 0 : x = 0 , x = π/3
k= 1 : x = 2π , x = π
k = 2 ; x = 5π/3
All the possible values of x such that [tex]\rm sin 2x = sin x\;and \; 0\leq x \leq 2\pi[/tex] are:[tex]\rm 0,\;\dfrac{\pi}{3},\;\pi,\;2\pi,\;and\; \dfrac{5\pi}{3}[/tex] and this can be evaluated by using the trignometry properties.
Given :
[tex]\rm sin 2x = sin x\;and \; 0\leq x \leq 2\pi[/tex]
All possible values of x can be determine by using trignometry properties. According to trignometry properties:
If [tex]\rm sin \theta = sin \alpha[/tex] than [tex]\rm (\theta = \alpha +2K\pi)\;\;\; or \;\; \; (\theta = \pi - \alpha +2K\pi)[/tex].
Now put the values of [tex]\theta[/tex] which is 2x and [tex]\alpha[/tex] which is x in above expression.
[tex]\rm sin2x = sin x \; \; \to \;\; (2x = x +2K\pi) \;\;or\;\; (2x = \pi-x+2K\pi )[/tex]
[tex]\Rightarrow \; x = 2K\pi \;\;or \;\; 3x=\pi(1+2K)[/tex]
[tex]\Rightarrow \; x = 2K\pi \;\;or \;\; x=\dfrac{\pi}{3}(1+2K)[/tex]
But the above solution is the general solution, the question ask the value of x between 0 and [tex]2\pi[/tex].
Therefore, at k = 0: [tex]x = 0 \;and \; x = \dfrac{\pi}{3}[/tex]
at k = 1: [tex]\rm x = 2\pi \; or \; x = \pi[/tex]
at k = 2: [tex]x = \dfrac{5\pi}{3}[/tex]
So, all the possible values of x such that [tex]\rm sin 2x = sin x\;and \; 0\leq x \leq 2\pi[/tex] are: [tex]\rm 0,\;\dfrac{\pi}{3},\;\pi,\;2\pi,\;and\; \dfrac{5\pi}{3}[/tex]
For more information, refer the link given below:
https://brainly.com/question/21286835
