Trigonometric Function 13.

Question

Find the solutions of the equation: sin(2x)=1/2 in the interval [0,2pi).

 

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Answer 1

Let   q = 2x

 

sin(q) = 1/2

 

q = π/6

 

q = π - (π/6) = 5π/6   --->  2nd quadrant

 

 

Plug in the values of q to solve for x.

 

2x = π/6             and             2x = 5π/6

 

Answer 2

sin(2x) = 1/2

 

2x = π/6 + 2kπ or 2x = 5π/6 + 2kπ, where k = 0, ±1, ±2, ...

 

So, x = π/12 + kπ, 5π/12 + kπ

 

If k = 0, then x = π/12, 5π/12

 

If k = 1, then x = 13π/12, 17π/12

 

Other values of k give values of x outside of the interval [0, 2π]

 

There are a total of 4 solutions in the interval [0, 2π]:

 

   π/12, 5π/12, 13π/12, 17π/12