Question

Pls need help now. ​

Answer 1

Answer:

  θ = {π/6, 5π/6} +2kπ . . . . for any integer k

Step-by-step explanation:

Multiplying by the product of the denominators, we can simplify this to ...

  sin(θ)² +(1+cos(θ))² = 4sin(θ)(1+cos(θ))

  sin(θ)² +1 +2cos(θ) +cos(θ)² = 4sin(θ)+4sin(θ)cos(θ)

  2 +2cos(θ) = 2sin(θ)(2 +2cos(θ)) . . . . show similar factors

  2(2sin(θ) -1)(1 +cos(θ)) = 0 . . . . subtract left side, complete the factoring

These factors are zero when ...

  2sin(θ) -1 = 0   ⇒   sin(θ) = 1/2   ⇒   θ = π/6, 5π/6

  1 +cos(θ) = 0 . . . . . extraneous solution; makes equation undefined

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Solutions are periodic with period 2π, so the complete solution set is ...

 θ = {π/6, 5π/6} +2kπ . . . . for any integer k