△=(2k+ 1)^2-4(k^2- 1)>； =0

Solution: k & gt=-5/4

According to Vieta's theorem:

Sinθ+cosθ=2k+ 1, and the two sides are divided equally to get: sin θ cos θ = 2k 2+2k.

Sinθ cos θ = k 2-1= 2k 2+2k, and the solution is k=- 1.

sinθcosθ=k^2- 1=0

sinθ+cosθ=2k+ 1=- 1

Sinθ=0, cosθ=- 1 or sinθ=- 1, cosθ =0.

The solution is: θ=2nπ+π or θ=2nπ-π/2, n ∈ z.