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We have,
∵ the general solution is
cosec θ = – √2
∵ sin(-θ) = -sin θ.
sin 9θ = sin θ
sin 9θ – sin θ = 0
Apply sin A sin B formula
sin 9θ – sin θ = 2 cos 5θ sin 4θ = 0
cos 5θ sin 4θ = 0
⟹ cos 5θ = 0(or) sin 4θ = 0
sin 2θ = cos 3θ
⟹ cos 3θ = sin2θ
⟹ either
tan θ + cot 2θ = 0
tan θ = – cot 2θ
⟹ cot 2θ = – tanθ
⟹ tan 2θ = – cotθ
tan 3θ = cotθ
tan 2θ tanθ = 1
⟹ tan 2θ = cotθ
tan mθ + cot nθ = 0
sin mθ sin nθ + cos mθ cos nθ = 0
cos (m – n)θ = 0
tan pθ = cotqθ
sin 2x + cos x = 0
2 sinx cosx + cosx = 0
cosx (2 sin +1) = 0
cosx = 0 or 2 sinx + 1 = 0
sin θ = tan θ
⟹ sinθ (cosθ – 1) = 0
⟹ either sinθ = 0 or cosθ – 1 = 0
⟹ θ = nπ, n ϵ z or cosθ = 1
⟹ cosθ = cos 0°
θ = 2mπ, m ϵ z
Thus,
θ = nπn ϵ z or θ = 2mπ, m ϵ z
cos (2x) = – sin(3x)
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