Solution of the equation sin 2x - sin x = 0 on the interval [0 , 2π) are :
x = { 0 , 1.05 , 3.14 , 5.24 }
Firstly , let us learn about trigonometry in mathematics.
Suppose the ΔABC is a right triangle and ∠A is 90°.
There are several trigonometric identities that need to be recalled, i.e.
[tex]cosec ~ A = \frac{1}{sin ~ A}[/tex]
[tex]sec ~ A = \frac{1}{cos ~ A}[/tex]
[tex]cot ~ A = \frac{1}{tan ~ A}[/tex]
[tex]tan ~ A = \frac{sin ~ A}{cos ~ A}[/tex]
Let us now tackle the problem!
[tex]\sin 2x - \sin x = 0[/tex]
[tex]2 \sin x \cos x - \sin x = 0[/tex]
[tex]\sin x (2 \cos x - 1) = 0[/tex]
[tex]\sin x = 0 ~ or ~ 2 \cos x - 1 = 0[/tex]
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.
Solution of the equation sin 2x - sin x = 0 on the interval [0 , 2π) are :
x = { 0 , 1.05 , 3.14 , 5.24 }
Grade: College
Subject: Mathematics
Chapter: Trigonometry
Keywords: Sine , Cosine , Tangent , Opposite , Adjacent , Hypotenuse
