find all solutions of the given equation in the interval [0,2pi);

Question

(4sin²x-1)sinx=0

Answer 1

We want to solve for x

 

sinx=0 and (4sin²x-1)=0   split into two equations

 

(2sinx-1)(2sinx+1)=0  factor the second (difference of two squares)

 

so now we have 3 factors whoas product is 0 so

 

sinx=1/2, sinx=-1/2 and sinx=0

 

taking the inverse of sinx=0 gives x=0,π as the solution for sinx=0

 

next we find the solution of sinx=1/2 or x=π/6, 5π/6,

There are two more solutions to sinx =-1/2 can you find them? When you do you will have found all the solutions to (4sin²x-1)*sinx = 0 in the interval [0,2π)

 

Jim

Answer 2

 

 

The function is sinx(4sin²x - 1) = 0. Therefore, either sinx = 0 or 4sin²x - 1 = 0 to find all the zeros of

this function. For sinx, we know that it equals 0 at 0, pi and 2pi, (0 degrees, 180 degrees and 360 degrees).

 

For the other function, we can solve by writing it as 4sin²x = 1. Divide both sides by 4, and we get

sin² x = 1/4.  So sinx = 1/2 by taking the square root of both sides. The arcsin of 1/2 = pi/6 radians, that is, 30 degrees.

 

Since the sine is positive in quadrants I and II, we take the equivalent angle by finding (180 - 30) degrees,

which equals 150 degrees or (pi - pi/6), finding the common denominator get 5pi/6.

 

Final answer is 0, pi/6, 5pi/6, pi, 2pi or 0, 30 degrees, 150 degrees, 180 degrees and 360 degrees.

 

Dorene O.