trigonometry

Question

find all solutions in the interval [0,2pi)
2sin^2 theta + sin theta = 0
Please help me. thank you.

Answer 1

Let x=θ.

 

 

2sin²x + sinx = 0

 

 

Factor out a sinx.

 

sinx(2sinx + 1) = 0

 

 

Set the factors equal to zero and solve for x.

 

sinx = 0             and              2sinx + 1 = 0

 

x = 0                                      sinx = -1/2

x = π                                          

x = 2π                                         x = -π/6

 

 

                                                 sine is negative in 3rd and 4th quadrants.

 

                                                   x = π + (π / 6)

                                                   x = (6π + π) / 6

                                                   x = 7π / 6

 

                                                  x = 2π - (π / 6)

                                                  x = (12π - π) / 6

                                                  x = 11π / 6

 

 

Your solutions are in bold.  There are 5 solutions.  Plug them into the original equation to verify.

Answer 2

2 sin² θ + sin θ = 0            factor out sin θ

sin θ (2 sin θ + 1) = 0              zero product rule

 

sin θ = 0

sin^-1 0 = θ

 

{0. π}

 

2 sin θ + 1 = 0

2 sin θ = -1

sin θ = -1/2

sin^-1 (-1/2) = θ

 

{7π/6, 11π/6}

Answer 3

2 sin²θ + sin θ = 0

Factoring, sin θ (2 sinθ + 1 ) = 0

→ sin θ = 0  ⇒  θ =0, π

or 2 sinθ + 1  = 0 ⇒ sin θ = -1/2

Since sinθ is negative, then θ must be a third or fourth quadrant angle.

                       ⇒ θ = 7π/6, 11π/6

 

Answer: {0, π, 7π/6, 11π/6}