Solving For x in the interval 0<x<2p

Question

Solve for x in the interval 0<x<2π (pi):

 

3sin²(2x+1)+2sin(2x+1)-1=0

 

Please show the steps! :)

Answer 1

So, you have sin²(2x + 1), sin(2x + 1), and a constant. What if you could treat sin(2x + 1) as a "symbol", like "a", temporarily? Substitute "a" for sin(2x + 1):

 

3a² + 2a - 1 = 0

 

Solve the quadratic for "a". I'll use completing the square:

 

3a² + 2a = 1, A = 3, divide both sides by A

a² + (2/3)a = 1/3, B = 2/3; B/2 = 1/3; B²/4 = 1/9, add B²/4 to both sides

a² + (2/3)a + 1/9 = 4/9, make the left side (a + B/2)²

(a + 1/3)² = 4/9, square root both sides

a + 1/3 = ±2/3

a + 1/3 = 2/3: a = 1/3

a + 1/3 = -2/3: a = -1

 

So "a" can equal 1/3 or -1. Turn "a" back into sin(2x + 1)!

 

sin(2x + 1) = 1/3

sin(2x + 1) = -1

 

The easiest way to figure out the values of x would be to graph:

 

y = sin(2x + 1)

y = 1/3 [-1, in the second equation]

 

and find where they intersect, 0 < x < 2π [~6.28].

 

I got x ~ 0.901, 2.812, 4.042, 5.953 [first equation], 1.856, 4.998 [second equation].