Question

Find all values of x such that sin 2x = sin x and 0 ≤ x ≤ 2π. (enter your answers as a comma-separated list.)

Answer 1

Hello : 
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

Answer 2

Answer:

Solution: $x=0,\dfrac{\pi}{3},\pi,\dfrac{5\pi}{3},2\pi$

Step-by-step explanation:

Given: $\sin 2x=\sin x$

Using trigonometry formula:

$\sin 2x=\sin x$

$2\sin x\cos x=\sin x$

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

$\sin x(2\cos x-1)=0$

$\sin x=0\text{ and }2\cos x-1=0$

$\sin x=0$

$x=0,\pi,2\pi$

$2\cos x=1$

$\cos x=\dfrac{1}{2}$

$x=\dfrac{\pi}{3},\dfrac{5\pi}{3}$

Hence, The solutions of the given equation are $x=0,\dfrac{\pi}{3},\pi,\dfrac{5\pi}{3},2\pi$