Question

（sinθ+3cosθ）（sinθ-cosθ）=-3

Answer 1

（sinθ+3cosθ）（sinθ-cosθ）=-3
sin²θ-3cos²θ+2sinθcosθ=-3
(1-cos2θ)/2-3(1+cos2θ)/2+sin2θ=-3
sin2θ-2cos2θ=-2
√5sin(2θ-arccos√5/5)=-2
sin(2θ-arccos√5/5)=-2√5/5
2θ-arccos√5/5=nπ-(-1)ⁿarcsin(2√5/5)
θ=[nπ-(-1)ⁿarccos√5/5-arccos√5/5]/2,n∈Z

追问

求tanθ，谢谢O(∩_∩)O谢谢

追答

sin²θ-3cos²θ+2sinθcosθ=-3
两边同时除以sin²θ+cos²θ=1
(sin²θ-3cos²θ+2sinθcosθ)/(sin²θ+cos²θ)=-3
(tan²θ-3+2tanθ)/(tan²θ+1)=-3 (上下同时除以cos²θ)
tan²θ-3+2tanθ=-3tan²θ-3
4tan²θ+2tanθ=0
tanθ=0,或tanθ=-1/2