2sin(2x)=sqrt3 between 0 and 2pi?

Question

The answers are pi/6, pi/3, 7pi/6, and 4pi/3. I think I understand how to get pi/6 and 7pi/6, but I'm not so sure about where pi/3 and 4pi/3 come from in relation to the problem. Any help?

Answer 1

Now over the interval we have

So,

So, the solutions are .

You have been asked about the solutions .

Of course, they are solutions of x because, if we consider the entire sine function we have

But these are not solutions over the interval .

Answer 2

The trigonometric equation is 2 sin(2x) = √3 and solve for x.

sin(2x) = √3/2

2x = sin^-1(√3/2)

The function sin(x) has a period of 2π, first find all solutions in the interval [0, 2π).

The function sin(x) is positive in first and second quadrant.

In First Quadrant, √3/2 = sin (π/3).

In Second Quadrant, √3/2 = sin (π/3) = sin (π - π/3) = sin (2π/3).

So the solutions are

2x = sin^-1(sin (π/3)) -------> 2x = π/3 ------> x = π/6.

2x = sin^-1(sin (2π/3)) -------> 2x = 2π/3 ------> x = π/3.

The solutions are 2x = π/3 and 2x = 2π/3.

Finally, add multiples of 2π to each of these solutions to get the general form

2x = π/3 + 2nπ and 2x = 2π/3 + 2nπ

x = π/6 + nπ and x = π/3 + nπ where n is integer.

The solutions are x = π/6 and x = π/3 in the interval [0, 2π).

Comments

The solutions are x = π/6 and x = π/3 in the interval [0, 2π).

The general forms of the solutions are x = π/6 + nπ and x = π/3 + nπ where n is integer.

If n = - 1 then

x = π/6 + nπ -----> x = π/6 - π -----> x = - 5π/6 and

x = π/3 + nπ -----> x = π/3 - π -----> x = - 2π/3.

If n = 0 then

x = π/6 + nπ -----> x = π/6 + 0 -----> x = π/6 and

x = π/3 + nπ -----> x = π/3 + 0 -----> x = π/3.

If n = 1 then

x = π/6 + nπ -----> x = π/6 + π -----> x = 7π/6 and

x = π/3 + nπ -----> x = π/3 + π -----> x = 4π/3.

The solutions are π/6, π/3, 7π/6 and 4π/3.