Solution of the equation sin 2x - sin x = 0 on the interval [0 , 2π) are :
x = { 0 , 1.05 , 3.14 , 5.24 }
<h3>Further explanation</h3>
Firstly , let us learn about trigonometry in mathematics.
Suppose the ΔABC is a right triangle and ∠A is 90°.
<h3>sin ∠A = opposite / hypotenuse</h3><h3>cos ∠A = adjacent / hypotenuse</h3><h3>tan ∠A = opposite / adjacent </h3>
There are several trigonometric identities that need to be recalled, i.e.
![cosec ~ A = \frac{1}{sin ~ A}](https://tex.z-dn.net/?f=cosec%20~%20A%20%3D%20%5Cfrac%7B1%7D%7Bsin%20~%20A%7D)
![sec ~ A = \frac{1}{cos ~ A}](https://tex.z-dn.net/?f=sec%20~%20A%20%3D%20%5Cfrac%7B1%7D%7Bcos%20~%20A%7D)
![cot ~ A = \frac{1}{tan ~ A}](https://tex.z-dn.net/?f=cot%20~%20A%20%3D%20%5Cfrac%7B1%7D%7Btan%20~%20A%7D)
![tan ~ A = \frac{sin ~ A}{cos ~ A}](https://tex.z-dn.net/?f=tan%20~%20A%20%3D%20%5Cfrac%7Bsin%20~%20A%7D%7Bcos%20~%20A%7D)
Let us now tackle the problem!
![\sin 2x - \sin x = 0](https://tex.z-dn.net/?f=%5Csin%202x%20-%20%5Csin%20x%20%3D%200)
![2 \sin x \cos x - \sin x = 0](https://tex.z-dn.net/?f=2%20%5Csin%20x%20%5Ccos%20x%20-%20%5Csin%20x%20%3D%200)
![\sin x (2 \cos x - 1) = 0](https://tex.z-dn.net/?f=%5Csin%20x%20%282%20%5Ccos%20x%20-%201%29%20%3D%200)
![\sin x = 0 ~ or ~ 2 \cos x - 1 = 0](https://tex.z-dn.net/?f=%5Csin%20x%20%3D%200%20~%20or%20~%202%20%5Ccos%20x%20-%201%20%3D%200)
If sin x = 0 , then for the interval [0 , 2π) → x = { 0 , 3.14 }
For 2 cos x - 1 = 0 :
2 cos x = 0 + 1
2 cos x = 1
cos x = ½
If cos x = ½ , then for the interval [0 , 2π) → x = { 1.05 , 5.24 }
If we draw a graph from the function above, it will look like the picture in the attachment.
<h2>Conclusion :</h2>
Solution of the equation sin 2x - sin x = 0 on the interval [0 , 2π) are :
x = { 0 , 1.05 , 3.14 , 5.24 }
<h3>Learn more</h3>
<h3>Answer details</h3>
Grade: College
Subject: Mathematics
Chapter: Trigonometry
Keywords: Sine , Cosine , Tangent , Opposite , Adjacent , Hypotenuse